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Exercise MCQ

Question 1

Which of the following is not a criterion for congruence of triangles?

(a) SSA

(b) SAS

(c) ASA

(d)SSSSolution 1

Correct option: (a)

SSA is not a criterion for congruence of triangles.Question 2

If AB = QR, BC = RP and CA = PQ, then which of the following holds?

(a) ∆ABC ≅ ∆PQR

(b) ∆CBA ≅ ∆PQR

(c) ∆CAB ≅ ∆PQR

(d) ∆BCA ≅ ∆PQRSolution 2

Correct option: (c)

Question 3

If ∆ABC ≅ ∆PQR then which of the following is not true?

(a) BC = PQ

(b) AC = PR

(c) BC = QR

(d) AB = PQSolution 3

Question 4

In Δ ABC, AB = AC and ∠B = 50°. Then, ∠A = ?

(a) 40° 

(b) 50° 

(c) 80° 

(d) 130° Solution 4

Correct option: (c)

In ΔABC,

AB = AC

⇒ ∠C = ∠B (angles opposite to equal sides are equal)

⇒ ∠C = 50° 

Now, ∠A + ∠B + ∠C = 180° 

⇒ ∠A + 50° + 50° = 180° 

⇒ ∠A = 80° Question 5

In Δ ABC, BC = AB and ∠B = 80°. Then, ∠A = ?

(a) 50° 

(b) 40° 

(c) 100° 

(d) 80° Solution 5

Correct option: (a)

In ΔABC,

BC = AB

⇒ ∠A = ∠C (angles opposite to equal sides are equal)

Now, ∠A + ∠B + ∠C = 180° 

⇒ ∠A + 80° + ∠A = 180° 

⇒ 2∠A = 100° 

⇒ ∠A = 50°  Question 6

In ΔABC, ∠C = ∠A, BC = 4 cm and AC = 5 cm. Then, AB = ?

(a) 4 cm

(b) 5 cm

(c) 8 cm

(d) 2.5 cmSolution 6

Correct option: (a)

In ΔABC,

∠C = ∠A

⇒ AB = BC (sides opposite to equal angles are equal)

⇒ AB = 4 cm Question 7

Two sides of a triangle are of length 4 cm and 2.5 cm. The length of the third side of the triangle cannot be

(a) 6 cm

(b) 6.5 cm

(c) 5.5 cm

(d) 6.3 cmSolution 7

Correct option: (b)

The sum of any two sides of a triangle is greater than the third side.

Since, 4 cm + 2.5 cm = 6.5 cm

The length of third side of a triangle cannot be 6.5 cm. Question 8

In ΔABC, if ∠C > ∠B, then

(a) BC > AC

(b) AB > AC

(c) AB < AC

(d) BC < ACSolution 8

Correct option: (b)

We know that in a triangle, the greater angle has the longer side opposite to it.

In ΔABC,

∠C > ∠B

⇒ AB >AC  Question 9

It is given that ∆ABC ≅ ∆FDE in which AB = 5 cm, ∠B = 40^o, ∠A = 80^o and FD = 5 cm. Then which of the following is true?

(a) ∠D = 60^o

(b) ∠E = 60^o

(c) ∠F = 60^o

(d) ∠D = 80^oSolution 9

Question 10

In ∆ABC, ∠A = 40^o and ∠B = 60^o. Then the longest side of ∆ABC is

(a) BC

(b) AC

(c) AB

(d) Cannot be determinedSolution 10

Question 11

In the given figure AB > AC. Then, which of the following is true?

(a) AB < AD

(b) AB = AD

(c) AB > AD

(d) Cannot be determined

Solution 11

Correct option: (c)

Question 12

In the given figure AB > AC. If BO and CO are the bisectors of ∠B and ∠C respectively, then

(a) OB = OC

(b) OB > OC

(c) OB < OC

Solution 12

Question 13

In the given figure, AB = AC and OB = OC. Then, ∠ABO : ∠ACO = ?

(a) 1 :1

(b) 2 : 1

(c) 1 :2

(d) None of these 

Solution 13

Question 14

If the altitudes from two vertices of a triangle to the opposite sides are equal, then the triangle is

(a) Equilateral

(b) Isosceles

(c) Scalene

(d) Right-angledSolution 14

Question 15

In ∆ABC and ∆DEF, it is given that AB = DE and BC = EF. In order that ∆ABC ≅ ∆DEF, we must have

(a) ∠A = ∠D

(b) ∠B = ∠E

(c) ∠C = ∠F

(d) None of these

Solution 15

Question 16

In ∆ABC and ∆DEF, it is given that ∠B = ∠E and ∠C = ∠F. In order that ∆ABC ≅ ∆DEF, we must have

(a) AB = DF

(b) AC = DE

(c) BC = EF

(d) ∠A = ∠D

Solution 16

Question 17

In ∆ABC and ∆PQR, it is given that AB = AC, ∠C = ∠P and ∠B = ∠Q. Then, the two triangles are

(a) Isosceles but not congruent

(b) Isosceles but congruent

(c) Congruent but not isosceles

(d) Neither congruent nor isosceles

Solution 17

Question 18

Which is true ?

(a) A triangle can have two right angles.

(b) A triangle can have two obtuse angles.

(c) A triangle can have two acute angles.

(d) An exterior angle of a triangle is less than either of the interior opposite angles.

Solution 18

Question 19

Fill in the blanks with

(a) (Sum of any two sides of a triangle)……(the third side)

(b) (Difference of any two sides of a triangle)…..(the third side)

(c) (Sum of three altitudes of a triangle) ……. (sum of its three sides

(d) (Sum of any two sides of a triangle)….. (twice the median to the 3^rd side)

(e) (Perimeter of a triangle)……(sum of its medians)Solution 19

Question 20

Fill in the blanks

(a) Each angle of an equilateral triangles measures …….

(b) Medians of an equilateral triangle are ……….

(c) In a right triangle the hypotenuse is the ….. side

(d) Drawing a ∆ABC with AB = 3cm, BC= 4 cm and CA = 7 cm is ……..Solution 20

Exercise Ex. 9B

Question 1(i)

Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.

5 cm, 4 cm, 9 cmSolution 1(i)

No, it is not possible to construct a triangle with lengths of its sides given because the sum of two sides, 5 cm and 4 cm, is not greater than the third side, 9 cm. Question 1(ii)

Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.

8 cm, 7 cm, 4 cmSolution 1(ii)

Yes, it is not possible to construct a triangle with lengths of its sides given because the sum of any two sides is greater than the third side. Question 1(iii)

Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.

10 cm, 5 cm, 6 cmSolution 1(iii)

Yes, it is not possible to construct a triangle with lengths of its sides given because the sum of any two sides is greater than the third side. Question 1(iv)

Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.

2.5 cm, 5 cm, 7 cmSolution 1(iv)

Yes, it is not possible to construct a triangle with lengths of its sides given because the sum of any two sides is greater than the third side. Question 1(v)

Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.

3 cm, 4 cm, 8 cmSolution 1(v)

No, it is not possible to construct a triangle with lengths of its sides given because the sum of two sides, 3 cm and 4 cm, is not greater than the third side, 8 cm. Question 2

In ΔABC, ∠A = 50° and ∠B = 60°. Determine the longest and shortest sides of the triangle.Solution 2

In ΔABC,

∠A + ∠B + ∠C = 180° 

⇒ 50° + 60° + ∠C = 180° 

⇒ ∠C = 70° 

Thus, we have

∠A < ∠B < ∠C

⇒ BC < AC < AB

Hence, the longest side is AB and the shortest side is BC. Question 3(iii)

In ΔABC, ∠A = 100° and ∠C = 50°. Which is its shortest side?Solution 3(iii)

In ΔABC,

∠A + ∠B + ∠C = 180° 

⇒ 100° + ∠B + 50° = 180° 

⇒ ∠B = 30° 

Thus, we have

∠B < ∠C < ∠A

⇒ AC < AB < BC

Hence, the shortest side is AC. Question 3(i)

In ABC, if A = 90^o, which is the longest side?Solution 3(i)

Question 3(ii)

In ABC, if A = B = 45^o, name the longest side.Solution 3(ii)

Question 4

In ABC, side AB is produced to D such that BD = BC. If B = 60^o and A = 70^o, prove that (i) AD > CD and (ii) AD > AC.

Solution 4

Question 5

In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

Solution 5

In ΔAOB,

∠B < ∠A

⇒ AO < BO ….(i)

In ΔCOD,

∠C < ∠D

⇒ DO < CO ….(ii)

Adding (i) and (ii),

AO + DO < BO + CO

⇒ AD < BC Question 6

AB and CD are respectively the smallest and largest sides of quadrilateral ABCD. Show that ∠A > ∠C and ∠B > ∠D.

Solution 6

Construction: Join AC and BD.

In ΔABC,

BC > AB

⇒ ∠BAC > ∠ACB ….(i)

In ΔACD,

CD > AD

⇒ ∠CAD > ∠ACD ….(ii)

Adding (i) and (ii), we get

∠BAC + ∠CAD > ∠ACB + ∠ACD

⇒ ∠A > ∠C 

In ΔADB,

AD > AB

⇒ ∠ABD > ∠ADB ….(iii)

In ΔBDC,

CD > BC

⇒ ∠CBD > ∠BDC ….(iv)

Adding (iii) and (iv), we get

∠ABD + ∠CBD > ∠ADB + ∠BDC

⇒ ∠B > ∠D Question 7

In a quadrilateral ABCD, show that

(AB + BC + CD + DA) > (AC + BD).Solution 7

In ΔABC,

AB + BC > AC  ….(i)

In ΔACD,

DA + CD > AC  ….(ii)

In ΔADB,

DA + AB > BD  ….(iii)

In ΔBDC,

BC + CD > BD  ….(iv)

Adding (i), (ii), (iii) and (iv), we get

AB + BC + DA + CD + DA + AB + BC + CD > AC + AC + BD + BD

⇒ 2(AB + BC + CD + DA) > 2(AC + BD)

⇒ AB + BC + CD + DA > AC + BD Question 8

In a quadrilateral ABCD, show that

(AB + BC + CD + DA) < 2(BD + AC).Solution 8

In ΔAOB,

AO + BO > AB  ….(i)

In ΔBOC,

BO + CO > BC  ….(ii)

In ΔCOD,

CO + DO > CD  ….(iii)

In ΔAOD,

DO + AO > DA  ….(iv)

Adding (i), (ii), (iii) and (iv), we get

AO + BO + BO + CO + CO + DO + DO + AO > AB + BC + CD + DA

⇒ 2(AO + CO) + 2(BO + DO) > AB + BC + CD + DA

⇒ 2AC + 2BD > AB + BC + CD + DA

⇒ 2(AC + BD) > AB + BC + CD + DA

⇒ AB + BC + CD + DA < 2(AC + BD) Question 9

In ABC, B = 35^o, C = 65^o and the bisector of BAC meets BC in X. Arrange AX, BX and CX in descending order.

Solution 9

Question 10

In the given figure, PQ > PR and QS and RS are the bisectors of ∠Q and ∠R respectively. Show that SQ > SR.

Solution 10

In ΔPQR,

PQ > PR

⇒ ∠PRQ > ∠PQR

⇒ ∠SRQ > ∠SQR

⇒ SQ > SRQuestion 11

D is any point on the side AC of ΔABC with AB = AC. Show that CD < BD.Solution 11

In ΔABC,

AB = AC

⇒ ∠ABC = ∠ACB ….(i)

Now, ∠ABC = ∠ABD + ∠DBC

⇒ ∠ABC > ∠DBC

⇒ ∠ACB > ∠DBC [From (i)]

⇒ ∠DCB > ∠DBC

⇒ BD > CD

i.e. CD < BD Question 12

Prove that in a triangle, other than an equilateral triangle, angle opposite to the longest side is greater than  of a right angle.Solution 12

Let PQR be the required triangle.

Let PR be the longest side.

Then, PR > PQ

⇒ ∠Q > ∠R ….(i)

Also, PR > QR

⇒ ∠Q > ∠P ….(ii)

Adding (i) and (ii), we get

2∠Q > ∠R + ∠P

⇒ 2∠Q + ∠Q > ∠P + ∠Q + ∠R (adding ∠Q to both sides)

⇒ 3∠Q > 180° 

⇒ ∠Q > 60° 

Question 13(i)

In the given figure, prove that CD + DA + AB > BC

Solution 13(i)

In ΔCDA,

CD + DA > AC ….(i)

In ΔABC,

AC + AB > BC ….(ii)

Adding (i) and (ii), we get

CD + DA + AC + AB > AC + BC

Subtracting AC from both sides, we get

CD + DA + AB > BC Question 13(ii)

In the given figure, prove that

CD + DA + AB + BC > 2AC.

Solution 13(ii)

In ΔCDA,

CD + DA > AC ….(i)

In ΔABC,

AB + BC > AC ….(ii)

Adding (i) and (ii), we get

CD + DA + AB + BC > AC + AC

⇒ CD + DA + AB + BC > 2AC Question 14(i)

If O is a point within ABC, show that:

AB + AC > OB + OCSolution 14(i)

Given : ABC is a triangle and O is appoint insideit.

To Prove : (i) AB+AC > OB +OCQuestion 14(ii)

If O is a point within ABC, show that:

AB + BC + CA > OA + OB + OCSolution 14(ii)

AB+BC+CA > OA+OB+OCQuestion 14(iii)

If O is a point within ABC, show that:

OA + OB + OC > (AB + BC + CA)Solution 14(iii)

OA+OB+OC> (AB+BC+CA)

Proof:

(i)InABC,

AB+AC>BC.(i)

And in , OBC,

OB+OC>BC.(ii)

Subtracting (i) from (i) we get

(AB+AC)-(OB+OC)> (BC-BC)

i.e.AB+AC>OB+OC

(ii)AB+AC> OB+OC[proved in (i)]

Similarly,AB+BC > OA+OC

AndAC+BC> OA +OB

Addingboth sides of these three inequalities, we get

(AB+AC) +(AC+BC) +(AB+BC)>OB+OC+OA+OB+OA+OC

i.e.2(AB+BC+AC)> 2(OA+OB+OC)

Therefore, we have

AB+BC+AC > OA+OB+OC

(iii)InOAB

OA+OB > AB(i)

InOBC,

OB+OC > BC(ii)

And, in OCA,

OC+OA>CA

Adding (i), (ii) and (iii)we get

(OA+OB) + (OB+OC)+(OC+OA)> AB+BC+CA

i.e2(OA+OB+OC) > AB+BC+CA

OA+OB+OC> (AB+BC+CA)Question 15

In the given figure, AD ⊥ BC and CD > BD. Show that AC > AB.

Solution 15

Construction: Mark a point S on BC such that BD = SD. Join AS.

In ΔADB and ΔADS,

BD = SD (by construction)

∠ADB = ∠ADS   (Each equal to 90°)

AD = AD (common)

∴ ΔADB ≅ ΔADS (by SAS congruence criterion)

⇒ AB = AS (c.p.c.t.)

Now, in ΔABS,

AB = AS

⇒ ∠ASB = ∠ABS ….(i)(angles opposite to equal sides are equal)

In ΔACS,

∠ASB > ∠ACS ….(ii)

From (i) and (ii), we have

∠ABS > ∠ACS

⇒ ∠ABC > ∠ACB

⇒ AC > ABQuestion 16

In the given figure, D is a point on side BC of a ΔABC and E is a point such that CD = DE. Prove that AB + AC > BE.

Solution 16

In ΔABC,

AB + AC > BC

⇒ AB + AC >BD + DC

⇒ AB + AC >BD + DE ….(i) [since CD = DE]

In ΔBED,

BD + DE > BE ….(ii)

From (i) and (ii), we have

AB + AC > BE

Exercise Ex. 9A

Question 1

In the given figure, AB ∥ CD and O is the midpoint of AD.

Show that (i) Δ AOB ≅ Δ DOC (ii) O is the midpoint of BC.

Solution 1

(i) In ΔAOB and ΔDOC,

∠BAO = ∠CDO (AB ∥ CD, alternate angles)

AO = DO (O is the mid-point of AD)

∠AOB = ∠DOC (vertically opposite angles)

∴ ΔAOB ≅ ΔDOC (by ASA congruence criterion)

(ii) Since ΔAOB ≅ ΔDOC,

BO = CO (c.p.c.t.)

⇒ O is the mid-point of BC.Question 2

In the given figure, AD and BC are equal perpendiculars to a line segment AB. Show that CD bisect AB.

Solution 2

In ΔAOD and ΔBOC,

∠AOD = ∠BOC (vertically opposite angles)

∠DAO = ∠CBO (Each 90°)

AD = BC (given)

∴ ΔAOD ≅ BOC (by AAS congruence criterion)

⇒ AO = BO (c.p.c.t.)

⇒ CD bisects AB. Question 3

In the given figure, two parallels lines l and m are intersected by two parallels lines p and q. Show that Δ ABC ≅ Δ CDA.

Solution 3

In ΔABC and ΔCDA

∠BAC = ∠DCA  (alternate interior angles for p ∥ q)

AC = CA  (common)

∠BCA = ∠DAC (alternate interior angles for l ∥ m)

∴ ΔABC ≅ ΔCDA (by ASA congruence rule)Question 4

AD is an altitude of an isosceles ΔABC in which AB = AC.

Show that (i) AD bisects BC, (ii) AD bisects ∠A.

Solution 4

(i) In ΔBAD and ΔCAD

∠ADB = ∠ADC  (Each 90° as AD is an altitude)

AB = AC (given)

AD = AD (common)

∴ ΔBAD ≅ ΔCAD (by RHS Congruence criterion)

⇒ BD = CD (c.p.c.t.)

Hence AD bisects BC.

(ii) Also, ∠BAD = ∠CAD (c.p.c.t.)

 Hence, AD bisects ∠A.Question 5

In the given figure, BE and CF are two equal altitudes of ΔABC.

Show that (i) ΔABE ≅ ΔACF, (ii) AB = AC.

Solution 5

(i) In ΔABE and ΔACF,

∠AEB = ∠AFC (Each 90°)

BE = CF (given)

∠BAE = ∠CAF (common ∠A)

∴ ΔABE ≅ ACF (by ASA congruence criterion)

(ii) Since ΔABE ≅ ΔACF,

AB = AC (c.p.c.t.)Question 6

ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at E, show that

(i) ΔABD ≅ ΔACD

(ii) ΔABE ≅ ΔACE

(iii) AE bisects ∠A as well as ∠D

(iv) AE is the perpendicular bisector of BC.

Solution 6

(i) In ΔABD and ΔACD,

AB = AC  (equal sides of isosceles ΔABC)

DB = DC  (equal sides of isosceles ΔDBC)

AD = AD (common)

∴ ΔABD ≅ ACD (by SSS congruence criterion)

(ii) Since ΔABD ≅ ΔACD,

∠BAD = ∠CAD (c.p.c.t.)

⇒ ∠BAE = ∠CAE ….(1)

Now, in ΔABE and ΔACE

AB = AC (equal sides of isosceles ΔABC)

∠BAE = ∠CAE [From (1)]

AE = AE (common) 

∴ ΔABE ≅ ACE (by SAS congruence criterion)

(iii) Since ΔABD ≅ ΔACD,

∠BAD = ∠CAD (c.p.c.t.)

⇒ ∠BAE = ∠CAE 

Thus, AE bisects ∠A.

In ΔBDE and ΔCDE,

BD = CD (equal sides of isosceles ΔABC)

BE = CE (c.p.c.t. since ΔABE ≅ ACE) 

DE = DE (common)

∴ ΔBDE ≅ CDE (by SSS congruence criterion)

⇒ ∠BDE = ∠CDE (c.p.c.t.)

Thus, DE bisects ∠D, i.e., AE bisects ∠D.

Hence, AE bisects ∠A as well as ∠D.

(iv) Since ΔBDE ≅ ΔCDE,

BE = CE and ∠BED = ∠CED (c.p.c.t.)

⇒ BE = CE and ∠BED = ∠CED = 90° (since ∠BED and ∠CED form a linear pair) 

⇒ DE is the perpendicular bisector of BC.

⇒ AE is the perpendicular bisector of BC.Question 7

In the given figure, if x = y and AB = CB, then prove that AE = CD.

Solution 7

Question 8

In the given figure, line l is the bisector of an angle ∠A and B is any point on l. If BP and BQ are perpendiculars from B to the arms of ∠A, show that

(i) ΔAPB ≅ ΔAQB

(ii) BP = BQ, i.e., B is equidistant from the arms of ∠A.

Solution 8

(i) In ΔAPB and ΔAQB,

∠APB = ∠AQC (Each 90°)

∠BAP = ∠BAQ (line l is the bisector of ∠A)

AB = AB (common)

∴ ΔAPB ≅ AQB (by AAS congruence criterion)

(ii) Since ΔAPB ≅ ΔAQB,

BP = BQ (c.p.c.t.) Question 9

ABCD is a quadrilateral such that diagonal AC bisect the angles ∠A and ∠C. Prove that AB = AD and CB = CD.Solution 9

In ΔABC and ΔADC,

∠BAC = ∠DAC (AC bisects ∠A)

AC = AC (common)

∠BCA = ∠DCA (AC bisects ∠C)

∴ ΔABC ≅ ADC (by ASA congruence criterion)

⇒ AB = AD and CB = CD (c.p.c.t.) Question 10

ΔABC is a right triangle right angled at A such that AB = AC and bisector of ∠C intersect the side AB at D. Prove that AC + AD = BC.Solution 10

Construction: Draw DE ⊥ BC.

In ΔDAC and ΔDEC,

∠DAC = ∠DEC (Each 90°)

∠DCA = ∠DCE (CD bisects ∠C)

CD = CD (common)

∴ ΔDAC ≅ ΔDEC (by AAS congruence criterion)

⇒ DA = DE (c.p.c.t.) ….(i)

and AC = EC (c.p.c.t.)  ….(ii)

Given, AB = AC

⇒ ∠B = ∠C (angles opposite to equal sides are equal)

In ΔABC, by angle sum property,

∠A + ∠B + ∠C = 180° 

⇒ 90° + ∠B + ∠B = 180° 

⇒ 2∠B = 90° 

⇒ ∠B = 45° 

In ΔBED,

∠BDE + ∠B = 90° (since ∠BED = 90°)

⇒ ∠BDE + 45° = 90° 

⇒ ∠BDE = 45° 

⇒ ∠BDE = ∠DBE = 45° 

⇒ DE = BE ….(iii)

From (i) and (iii),

DA = DE = BE ….(iv) 

Now, BC = BE + EC

⇒ BC = DA + AC [From (ii) and (iv)

⇒ AC + AD = BCQuestion 11

In the given figure, OA = OB and OP = OQ. Prove that (i) PX = QX (ii) AX = BX.

Solution 11

Question 12

In the given figure, ABC is an equilateral triangle; PQ || AC and AC is produced to R such that CR = BP. Prove that QR bisects PC.

Solution 12

Question 13

In the given figure, ABCD is a quadrilateral in which AB || DC and P is the midpoint BC. On producing, AP and DC meet at Q. Prove that (i) AB = CQ, (ii) DQ = DC + AB.

Solution 13

Question 14

In the given figure, ABCD is square and P is a point inside it such that PB = PD. Prove that CPA is a straight line.

Solution 14

Question 15

In the given figure, O is a point in the interior of square ABCD such that ΔOAB is an equilateral triangle. Show that ΔOCD is an isosceles triangle.

Solution 15

ΔOAB is an equilateral triangle.

⇒ ∠OAB = ∠OBA = AOB = 60° 

ABCD is a square.

⇒ ∠A = ∠B = ∠C = ∠D = 90° 

Now, ∠A = ∠DAO + ∠OAB

⇒ 90° = ∠DAO + 60° 

⇒ ∠DAO = 90° – 60° = 30° 

Similarly, ∠CBO = 30° 

In ΔOAD and ΔOBC,

AD = BC  (sides of a square ABCD)

∠DAO = ∠CBO = 30° 

OA = OB (sides of an equilateral ΔOAB)

∴ ΔOAD ≅ ΔOBC (by SAS congruence criterion)

⇒ OD = OC (c.p.c.t.)

Hence, ΔOCD is an isosceles triangle.Question 16

In the adjoining figure, X and Y are respectively two points on equal sides AB and AC of ABC such that AX = AY. Prove that CX = BY.

Solution 16

Question 17

In ABC, D is the midpoint of BC. If DL AB and DM AC such that DL = DM, prove that AB = AC.

Solution 17

Question 18

In ABC, AB = AC and the bisectors of _B and C meet at a point O. Prove that BO = CO and the ray AO is the bisector A.

Solution 18

Question 19

The line segments joining the midpoints M and N of parallel sides AB and DC respectively of a trapezium ABCD is perpendicular to both the sides AB and DC. Prove that AD = BC.Solution 19

Construction: Join AN and BN.

In ΔANM and ΔBNM

AM = BM (M is the mid-point of AB)

∠AMN = ∠BMN (Each 90°)

MN = MN (common)

∴ ΔANM ≅ ΔBNM (by SAS congruence criterion)

⇒ AN = BN (c.p.c.t.) ….(i)

And, ∠ANM = ∠BNM (c.p.c.t.)

⇒ 90° – ∠ANM = 90° – ∠BNM

⇒ ∠AND = ∠BNC ….(ii) 

In ΔAND and DBNC,

AN = BN [From (i)]

∠AND = ∠BNC [From (ii)]

DN = CN (N is the mid-point of DC)

∴ ΔAND ≅ ΔBNC (by SAS congruence criterion)

⇒ AD = BC (c.p.c.t.)Question 20

The bisectors of ∠B and ∠C of an isosceles triangle with AB = AC intersect each other at a point O. BO is produced to meet AC at a point M. Prove that ∠MOC = ∠ABC.Solution 20

In ΔABC, AB = AC

⇒ ∠ABC = ∠ACB

⇒ ∠OBC = ∠OCB ….(i)

Now, by exterior angle property,

∠MOC = ∠OBC + ∠OCB

⇒ ∠MOC = 2∠OBC [From (i)]

⇒ ∠MOC = ∠ABC (OB is the bisector of ∠ABC)  Question 21

The bisectors of ∠B and ∠C of an isosceles ΔABC with AB = AC intersect each other at a point O. Show that the exterior angle adjacent to ∠ABC is equal to ∠BOC.Solution 21

In ΔABC, AB = AC

⇒ ∠ABC = ∠ACB

⇒ ∠OBC = ∠OCB ….(i)

In ΔBOC, by angle sum property,

∠BOC + ∠OBC + ∠OCB = 180° 

⇒ ∠BOC + 2∠OBC = 180° [From (i)]

⇒ ∠BOC + ∠ABC = 180° 

⇒ ∠BOC + (180° – ∠ABP) = 180° (∠ABC and ∠ABP form a linear pair)

⇒ ∠BOC + 180° – ∠ABP = 180° 

⇒ ∠BOC – ∠ABP = 0

⇒ ∠BOC = ∠ABP Question 22

P is a point on the bisector of ∠ABC. If the line through P, parallel to BA meets BC at Q, prove that ΔBPQ is an isosceles triangle.Solution 22

AB ∥ PQ and BP is a transversal.

⇒ ∠ABP = ∠BPQ (alternate angles) ….(i)

BP is the bisector of ∠ABC.

⇒ ∠ABP = ∠PBC

⇒ ∠ABP = ∠PBQ ….(ii)

From (i) and (ii), we have

∠BPQ = ∠PBQ

⇒ PQ = BQ (sides opposite to equal angles are equal)

⇒ ΔBPQ is an isosceles triangle.Question 23

The image of an object placed at a point A before a plane mirror LM is seen at the point B by an observer at D, as shown in the figure. Prove that the image is as far behind the mirror as the object is in front of the mirror.

Solution 23

To prove that the image is as far behind the mirror as the object is in front of the mirror, we need to prove that AT = BT. 

We know that angle of incidence = angle of reflection.

⇒ ∠ACN = ∠DCN ….(i)

AB ∥ CN and AC is the transversal.

⇒ ∠TAC = ∠ACN (alternate angles) ….(ii)

Also, AB ∥ CN and BD is the transversal.

⇒ ∠TBC = ∠DCN (corresponding angles) ….(iii)

From (i), (ii) and (iii),

∠TAC = ∠TBC ….(iv)

In ΔACT and ΔBCT,

∠TAC = ∠TBC [From (iv)]

∠ATC = ∠BTC (Each 90°)

CT = CT (common)

∴ ΔACT ≅ ΔBCT (by AAS congruence criterion)

⇒ AT = BT (c.p.c.t.)Question 24

In the adjoining figure, explain how one can find the breadth of the river without crossing it.

Solution 24

_{Let AB be the breadth of a river. Now take a point M on that bank of the river where point B is situated. Through M draw a perpendicular and take point N on it such that point, A, O and N lie on a straight line where point O is the mid point of BM.}

Question 25

In a ΔABC, D is the midpoint of side AC such that BD = . Show that ∠ABC is a right angle.Solution 25

D is the mid-point of AC.

⇒ AD = CD =  

Given, BD =  

⇒ AD = CD = BD 

Consider AD = BD

⇒ ∠BAD = ∠ABD (i)(angles opposite to equal sides are equal)

Consider CD = BD

⇒ ∠BCD = ∠CBD (ii)(angles opposite to equal sides are equal)

In ΔABC, by angle sum property,

∠ABC + ∠BAC + ∠BCA = 180° 

⇒ ∠ABC + ∠BAD + ∠BCD = 180° 

⇒ ∠ABC + ∠ABD + ∠CBD = 180° [From (i) and (ii)]

⇒ ∠ABC + ∠ABC = 180° 

⇒ 2∠ABC = 180° 

⇒ ∠ABC = 90° 

Hence, ∠ABC is a right angle.Question 26

“If two sides and an angle of one triangle are equal to two sides and an angle of another triangle then the two triangles must be congruent.” Is the statement true? Why?Solution 26

The given statement is not true.

Two triangles are congruent if two sides and the included angle of one triangle are equal to corresponding two sides and the included angle of another triangle. Question 27

“If two angles and a side of one triangle are equal to two angles and a side of another triangle then the two triangles must be congruent.” Is the statement true? Why?Solution 27

The given statement is not true.

Two triangles are congruent if two angles and the included side of one triangle are equal to corresponding two angles and the included angle of another triangle. 

Exercise MCQ

Question 1

If (x + 1) is a factor of the polynomial (2x² + kx) then the value of k is

(a) -2

(b) -3

(c) 2

(d) 3Solution 1

Correct option: (c)

Let p(x) = 2x² + kx

Since (x + 1) is a factor of p(x),

P(-1) = 0

⇒ 2(-1)² + k(-1) = 0

⇒ 2 – k = 0

⇒ k = 2Question 2

The value of (249)² – (248)² is

(a) 1²

(b) 477

(c) 487

(d) 497Solution 2

Correct option: (d)

(249)² – (248)²

= (249 + 248)(249 – 248)

= 497 × 1

= 497Question 3

Solution 3

Question 4

If a + b + c = 0, then a³ + b³ + c³= ?

(a) 0

(b) abc 

(c) 2abc

(d)3abcSolution 4

Question 5

(a) 0

(b) 

(c) 

(d) Solution 5

Correct option: (c)

Question 6

The coefficient of x in the expansion of (x + 3)³ is

(a) 1

(b) 9

(c) 18

(d)27Solution 6

Question 7

Which of the following is a factor of (x + y)³ – (x³ + y³)?

(a) x² + y² + 2xy

(b) x² + y² – xy

(c) xy²

(d) 3xySolution 7

Correct option: (d)

(x + y)³ – (x³ + y³)

= (x + y)³ – [(x + y)(x² – xy + y²)]

= (x + y)[(x + y)² – (x² – xy + y²)]

= (x + y)[x² + y² + 2xy – x² + xy – y²]

= (x + y)(3xy) Question 8

One of the factors of (25x² – 1) + (1 + 5x)² is

(a) 5 +x

(b) 5 – x

(c) 5x – 1

(d) 10xSolution 8

Correct option: (d)

(25x² – 1) + (1 + 5x)²

= [(5x)² – (1)²] + (1 + 5x)²

= (5x – 1)(5x + 1) + (1 + 5x)²

= (1 + 5x)[(5x – 1) + (1 + 5x)]

= (1 + 5x)(5x – 1 + 1 + 5x)

= (1 + 5x)(10x)Question 9

If (x + 5) is a factor of p(x) = x³ – 20x + 5k then k = ?

(a) -5

(b) 5

(c) 3

(d) -3Solution 9

Correct option: (b)

Since (x + 5) is a factor of p(x) = x³ – 20x + 5k,

p(-5) = 0

⇒ (-5)³ – 20(-5) + 5k = 0

⇒ -125 + 100 + 5k = 0

⇒ -25 + 5k = 0

⇒ 5k = 25

⇒ k = 5 Question 10

If (x + 2) and (x – 1) are factors of (x³ + 10x² + mx + n), then

(a) m = 5, n = -3

(b) m = 7, n = -18

(c) m = 17, n = -8

(d)m = 23, n = -19Solution 10

Question 11

104 ⨯ 96 =?

(a) 9894

(b) 9984

(c) 9684

(d)9884Solution 11

Question 12

305 ⨯ 308 = ?

(a) 94940

(b) 93840

(c) 93940

(d)94840Solution 12

Question 13

207 ⨯ 193 = ?

(a) 39851

(b) 39951

(c) 39961

(d)38951Solution 13

Question 14

4a² + b² + 4ab + 8a + 4b + 4 =?

(a) (2a + b + 2)²

(b) (2a – b + 2)²

(c) (a + 2b + 2)²

(d)None of theseSolution 14

Question 15

(x² – 4x – 21) =?

(a) (x – 7)(x – 3)

(b) (x + 7)(x – 3)

(c) (x – 7)(x + 3)

(d)none of theseSolution 15

Question 16

(4x² + 4x -3) = ?

(a) (2x – 1)(2x – 3)

(b) (2x + 1)(2x – 3)

(c) (2x + 3)(2x – 1)

(d)none of theseSolution 16

Question 17

6x² + 17x + 5 =?

(a) (2x +3)(3x + 5)

(b) (2x + 5)(3x + 1)

(c) (6x + 5)(x + 1)

(d)none of theseSolution 17

Question 18

(x + 1) is a factor of the polynomial

(a) x³ – 2x² + x + 2

(b) x³ + 2x² + x – 2

(c) x³ + 2x² – x – 2

(d)x³ + 2x² – x + 2Solution 18

Question 19

3x³ + 2x² + 3x + 2 =?

(a) (3x – 2)(x² – 1)

(b) (3x – 2)(x² + 1)

(c) (3x + 2)(x² – 1)

(d)(3x + 2)(x² + 1)Solution 19

Question 20

(a) 1

(b) 0

(c) -1

(d)3Solution 20

Question 21

If x + y + z =9 and xy + yz + zx = 23, then the value of (x³ + y³ + z³ – 3xyz) = ?

(a) 108

(b) 207

(c) 669

(d)729Solution 21

Question 22

then (a³ – b³) = ?

(a) -3

(b) -2

(c) -1

(d) 0Solution 22

Correct option: (d)

Exercise Ex. 3C

Question 1

Factorise:

x² + 11x + 30Solution 1

x² + 11x + 30

= x² + 6x + 5x + 30

= x (x + 6) + 5 (x + 6)

= (x + 6) (x + 5).Question 2

Factorise:

x² + 18x + 32Solution 2

x² + 18x + 32

= x² + 16x + 2x + 32

= x (x + 16) + 2 (x + 16)

= (x + 16) (x + 2).Question 3

Factorise:

x² + 20x – 69Solution 3

x² + 20x – 69

= x² + 23x – 3x – 69

= x(x + 23) – 3(x + 23)

= (x + 23)(x – 3) Question 4

Factorise:

x² + 19x – 150Solution 4

x² + 19x – 150

= x² + 25x – 6x – 150

= x(x + 25) – 6(x + 25)

= (x + 25)(x – 6) Question 5

Factorise:

x² + 7x – 98Solution 5

x² + 7x – 98

= x² + 14x – 7x – 98

= x(x + 14) – 7(x + 14)

= (x + 14)(x – 7) Question 6

Factorise:

Solution 6

Question 7

Factorise:

x² – 21x + 90Solution 7

x² – 21x + 90

= x² – 6x – 15x + 90

= x(x – 6) – 15(x – 6)

= (x – 6)(x – 15) Question 8

Factorise:

x² – 22x + 120Solution 8

x² – 22x + 120

= x² – 10x – 12x + 120

= x(x – 10) – 12(x – 10)

= (x – 10)(x – 12) Question 9

Factorise:

x² – 4x + 3Solution 9

x² – 4x + 3

= x² – 3x – x + 3

= x(x – 3) – 1(x – 3)

= (x – 3)(x – 1) Question 10

Factorise:

Solution 10

Question 11

Factorise:

Solution 11

Question 12

Factorise:

Solution 12

Question 13

Factorise:

Solution 13

Question 14

Factorise:

x² – 24x – 180Solution 14

x² – 24x – 180

= x² – 30x + 6x – 180

= x(x – 30) + 6(x – 30)

= (x – 30)(x + 6) Question 15

Factorise:

z² – 32z – 105Solution 15

z² – 32z – 105

= z² – 35z + 3z – 105

= z (z – 35) + 3 (z – 35)

= (z – 35) (z + 3)Question 16

Factorise:

x² – 11x – 80Solution 16

x² – 11x – 80

= x² – 16x + 5x – 80

= x (x – 16) + 5 (x – 16)

= (x – 16) (x + 5).Question 17

Factorise:

6 – x – x²Solution 17

6 – x – x²

= 6 + 2x – 3x – x²

= 2(3 + x) – x (3 + x)

= (3 + x) (2 – x).Question 18

Factorise:

Solution 18

Question 19

Factorise:

40 + 3x – x²Solution 19

40 + 3x – x²

= 40 + 8x – 5x – x²

= 8 (5 + x) -x (5 + x)

= (5 + x) (8 – x).Question 20

Factorise:

x² – 26x + 133Solution 20

x² – 26x + 133

= x² – 19x – 7x + 133

= x(x – 19) – 7(x – 19)

= (x – 19)(x – 7) Question 21

Factorise:

Solution 21

Question 22

Factorise:

Solution 22

Question 23

Factorise:

Solution 23

Question 24

Factorise:

Solution 24

Question 25

Factorise:

x² – x – 156Solution 25

x² – x – 156

= x² – 13x + 12x – 156

= x (x – 13) + 12 (x – 13)

= (x – 13) (x + 12).Question 27

Factorise:

9x² + 18x + 8Solution 27

9x² + 18x + 8

= 9x² + 12x + 6x + 8

= 3x (3x+ 4) +2 (3x + 4)

= (3x + 4) (3x + 2).Question 28

Factorise:

6x² + 17x + 12Solution 28

6x² + 17x + 12

= 6x² + 9x + 8x + 12

= 3x (2x + 3) + 4(2x + 3)

= (2x + 3) (3x + 4).Question 29

Factorise:

18x² + 3x – 10Solution 29

18x² + 3x – 10

= 18x² – 12x + 15x – 10

= 6x (3x – 2) + 5 (3x – 2)

= (6x + 5) (3x – 2).Question 30

Factorise:

2x² + 11x – 21Solution 30

2x² + 11x – 21

= 2x² + 14x – 3x – 21

= 2x (x + 7) – 3 (x + 7)

= (x + 7) (2x – 3).Question 31

Factorise:

15x² + 2x – 8Solution 31

15x² + 2x – 8

= 15x² – 10x + 12x – 8

= 5x (3x – 2) + 4 (3x – 2)

= (3x – 2) (5x + 4).Question 32

Factorise:

21x² + 5x – 6Solution 32

21x² + 5x – 6

= 21x² + 14x – 9x – 6

= 7x(3x + 2) – 3(3x + 2)

= (3x + 2)(7x – 3) Question 33

Factorise:

24x² – 41x + 12Solution 33

24x² – 41x + 12

= 24x² – 32x – 9x + 12

= 8x (3x – 4) – 3 (3x – 4)

= (3x – 4) (8x – 3).Question 34

Factorise:

3x² – 14x + 8Solution 34

3x² – 14x + 8

= 3x² – 12x – 2x +8

= 3x (x – 4) – 2(x – 4)

= (x – 4) (3x – 2).Question 35

Factorise:

2x² + 3x – 90Solution 35

2x² + 3x – 90

= 2x² – 12x + 15x – 90

= 2x (x – 6) + 15 (x – 6)

= (x – 6) (2x + 15).Question 37

Factorise:

Solution 37

Question 38

Factorise:

Solution 38

Question 39

Factorise:

Solution 39

Question 40

Factorise:

Solution 40

Question 41

Factorise:

Solution 41

Question 42

Factorise:

Solution 42

Question 43

Factorise:

Solution 43

Question 44

Factorise:

15x² – x – 28Solution 44

15x² – x – 28

= 15x² + 20x – 21x – 28

= 5x (3x + 4) – 7 (3x + 4)

= (3x + 4) (5x – 7).Question 45

Factorise:

6x² – 5x – 21Solution 45

6x² – 5x – 21

= 6x² + 9x – 14x – 21

= 3x (2x + 3) – 7 (2x + 3)

= (3x – 7) (2x + 3).Question 46

Factorise:

2x² – 7x – 15Solution 46

2x² – 7x – 15

= 2x² – 10x + 3x – 15

= 2x (x – 5) + 3 (x – 5)

= (x – 5) (2x + 3).Question 47

Factorise:

5x² – 16x – 21Solution 47

5x² – 16x – 21

= 5x² + 5x – 21x – 21

= 5x (x + 1) -21 (x + 1)

= (x + 1) (5x – 21).Question 48

Factorise:

6x² – 11x – 35Solution 48

6x² – 11x – 35

= 6x² – 21x + 10x – 35

= 3x(2x – 7) + 5(2x – 7)

= (2x – 7)(3x + 5) Question 49

Factorise:

9x² – 3x – 20Solution 49

9x² – 3x – 20

= 9x² – 15x + 12x – 20

= 3x(3x – 5) + 4(3x – 5)

= (3x – 5)(3x + 4) Question 50

Factorise:

10x² – 9x – 7Solution 50

10x² – 9x – 7

= 10x² + 5x – 14x – 7

= 5x (2x + 1) – 7 (2x+ 1)

= (2x + 1) (5x – 7).Question 51

Factorise:

x² – 2x + Solution 51

x² – 2x + 

Question 52

Factorise:

Solution 52

Question 53

Factorise:

Solution 53

Question 54

Factorise:

Solution 54

Question 55

Factorise:

Solution 55

Question 56

Factorise:

Solution 56

Question 57

Factorise:

Solution 57

Question 58

Factorise:

Solution 58

Question 59

Factorise:

Solution 59

Letx + y = z

Then, 2 (x+ y)² – 9 (x + y) – 5

Now, replacing z by (x + y), we get

Question 60

Factorise:

9 (2a – b)² – 4 (2a – b) – 13Solution 60

Let2a – b = c

Then,9 (2a – b)² – 4 (2a – b) -13

Now, replacing c by (2a – b) , we get

9 (2a – b)² – 4 (2a – b) – 13

Question 61

Factorise:

7 (x – 2y)² – 25 (x – 2y) + 12Solution 61

Let x – 2y = z

Then, 7 (x – 2y)² – 25 (x – 2y) + 12

Now replace z by (x – 2y), we get

7 (x – 2y)² – 25 (x – 2y) + 12

Question 62

Factorise:

Solution 62

Question 63

Factorise:

Solution 63

Question 64

Factorise:

(a + 2b)² + 101(a + 2b) + 100Solution 64

Given equation: (a + 2b)² + 101(a + 2b) + 100

Let (a + 2b) = x

Then, we have

x² + 101x + 100

= x² + 100x + x + 100

= x(x + 100) + 1(x + 100)

= (x + 100)(x + 1)

= (a + 2b + 100)(a + 2b + 1) Question 65

Factorise:

Solution 65

Let x² = y

Then, 4x⁴ + 7x² – 2

Now replacing y by x², we get

Question 66

Evaluate {(999)² – 1}.Solution 66

{(999)² – 1}

= {(999)² – (1)²}

= {(999 – 1)(999 + 1)}

= 998 × 1000

= 998000 

Exercise Ex. 3D

Question 1(i)

Expand:

(a + 2b + 5c)²Solution 1(i)

We know:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

(i) (a + 2b + 5c)²

= (a)² + (2b)² + (5c)² + 2(a) (2b) + 2 (2b) (5c) + 2(5c) (a)

= a² + 4b² + 25c² + 4ab + 20bc + 10acQuestion 1(ii)

Expand:

(2a – b + c)²Solution 1(ii)

We know:

(2a – b + c)²

= (2a)² + (-b)² + (c)² + 2 (2a) (-b) + 2(-b) (c) + 2 (c) (2a)

= 4a² + b² + c² – 4ab – 2bc + 4ac.Question 1(iii)

Expand:

(a – 2b – 3c)²Solution 1(iii)

We know:

(a – 2b – 3c)²

= (a)² + (-2b)² + (-3c)²+ 2(a) (-2b) + 2(-2b) (-3c) + 2 (-3c) (a)

= a² + 4b² + 9c² – 4ab + 12bc – 6ac.Question 2(i)

Expand:

(2a – 5b – 7c)²Solution 2(i)

We know:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

(2a – 5b – 7c)²

= (2a)² + (-5b)² + (-7c)² + 2 (2a) (-5b) + 2 (-5b) (-7c) + 2 (-7c) (2a)

= 4a² + 25b² + 49c² – 20ab + 70bc – 28ac.Question 2(ii)

Expand:

(ii) (-3a + 4b – 5c)²Solution 2(ii)

(-3a + 4b – 5c)²

= (-3a)² + (4b)² + (-5c)² + 2 (-3a) (4b) + 2 (4b) (-5c) + 2 (-5c) (-3a)

= 9a² + 16b² + 25c² – 24ab – 40bc + 30ac.Question 2(iii)

Expand:

Solution 2(iii)

= Question 3

Factorise: 4x² + 9y² + 16z² + 12xy – 24yz – 16xz.Solution 3

4x² + 9y² + 16z² + 12xy – 24yz – 16xz

= (2x)² + (3y)² + (-4z)² + 2 (2x) (3y) + 2(3y) (-4z) + 2 (-4z) (2x)

= (2x + 3y – 4z)²Question 4

Factorise: 9x² + 16y² + 4z² – 24xy + 16yz – 12xzSolution 4

9x² + 16y² + 4z² – 24xy + 16yz – 12xz

= (-3x)² + (4y)² + (2z)² + 2 (-3x) (4y) + 2 (4y) (2z) + 2 (2z) (-3x)

= (-3x + 4y + 2z)².Question 5

Factorise: 25x² + 4y² + 9z² – 20xy – 12yz + 30xz.Solution 5

25x² + 4y² + 9z² – 20xy – 12yz + 30xz

= (5x)² + (-2y)² + (3z)² + 2(5x) (-2y) + 2(-2y) (3z) + 2(3z) (5x)

= (5x – 2y + 3z)²Question 6

16x² + 4y² + 9z² – 16xy – 12yz + 24xzSolution 6

16x² + 4y² + 9z² – 16xy – 12yz + 24xz

= (4x)² + (-2y)² + (3z)² + 2(4x)(-2y) + 2(-2y)(3z) + 2(4x)(3z)

= [4x + (-2y) + 3z]²

= (4x – 2y + 3z)²

= (4x – 2y + 3z)(4x – 2y + 3z) Question 7(i)

Evaluate

(99)²Solution 7(i)

(99)²

= (100 – 1)²

= (100)² – 2(100) (1) + (1)²

= 10000 – 200 + 1

= 9801.Question 7(ii)

Evaluate

(995)²Solution 7(ii)

(995)²

= (1000 – 5)²

= (1000)² + (5)² – 2(1000)(5)

= 1000000 + 25 – 10000

= 990025 Question 7(iii)

Evaluate

(107)²Solution 7(iii)

(107)²

= (100 + 7)²

= (100)² + (7)² + 2(100)(7)

= 10000 + 49 + 1400

= 11449 

Exercise Ex. 3E

Question 1(i)

Expand:

(3x + 2)³Solution 1(i)

(3x + 2)³

= (3x)³ + (2)³ + 3 3x 2 (3x + 2)

= 27x³ + 8 + 18x (3x + 2)

= 27x³ + 8 + 54x² + 36x.Question 1(ii)

Expand:

Solution 1(ii)

Question 1(iii)

Expand

Solution 1(iii)

Question 2(i)

Expand

(5a – 3b)³Solution 2(i)

(5a – 3b)³

= (5a)³ – (3b)³ – 3(5a)(3b)(5a – 3b)

= 125a³ – 27b³ – 45ab(5a – 3b)

= 125a³ – 27b³ – 225a²b + 135ab² Question 2(ii)

Expand

Solution 2(ii)

Question 2(iii)

Expand

Solution 2(iii)

Question 3

Factorise 

8a³ + 27b³ + 36a²b + 54ab²Solution 3

8a³ + 27b³ + 36a²b + 54ab²

= (2a)³ + (3b)³ + 3 × (2a)² × (3b) + 3 × (2a) × (3b)²

= (2a + 3b)³

= (2a + 3b)(2a + 3b)(2a + 3b) Question 4

Factorise 

64a³ – 27b³ – 144a²b + 108ab²Solution 4

64a³ – 27b³ – 144a²b + 108ab²

= (4a)³ – (3b)³ – 3 × (4a)² × (3b) + 3 × (4a) × (3b)²

= (4a – 3b)³

= (4a – 3b)(4a – 3b)(4a – 3b) Question 5

Factorise 

Solution 5

Question 6

Factorise 

125x³ – 27y³ – 225x²y + 135xy²Solution 6

125x³ – 27y³ – 225x²y + 135xy²

= (5x)³ – (3y)³ – 3 × (5x)² × (3y) + 3 × (5x) × (3y)²

= (5x – 3y)³

= (5x – 3y)(5x – 3y)(5x – 3y)  Question 7

Factorise 

a³x³ – 3a²bx² + 3ab²x – b³Solution 7

a³x³ – 3a²bx² + 3ab²x – b³

= (ax)³ – 3 × (ax)² × b + 3 × (ax) × (b)² – (b)³

= (ax – b)³

= (ax – b)(ax – b)(ax – b)  Question 8

Factorise 

Solution 8

Question 9

Factorise 

a³ – 12a(a – 4) – 64Solution 9

a³ – 12a(a – 4) – 64

= a³ – 3 × a × 4(a – 4) – (4)³

= (a – 4)³

= (a – 4)(a – 4)(a – 4) Question 10(i)

Evaluate

(103)³Solution 10(i)

(103)³

= (100 + 3)³

= (100)³ + (3)³ + 3 × 100 × 3 × (100 + 3)

= 1000000 + 27 + 900(103)

= 1000000 + 27 + 92700

= 1092727 Question 10(ii)

Evaluate

(99)³Solution 10(ii)

(99)³

= (100 – 1)³

= (100)³ – (1)³ – 3 × 100 × 1 × (100 – 1)

= 1000000 – 1 – 300(99)

= 1000000 – 1 – 29700

= 970299  

Exercise Ex. 3F

Question 1

Factorise:

x³ + 27Solution 1

x³ + 27

= x³ + 3³

= (x + 3) (x² – 3x + 9)

Question 2

Factorise 

27a³ + 64b³Solution 2

27a³ + 64b³

= (3a)³ + (4b)³

= (3a + 4b)[(3a)² – (3a)(4b) + (4b)²]

= (3a + 4b)(9a² – 12ab + 16b²) Question 3

Factorise:

125a³ + Solution 3

125a³ + 

We know that 

Let us rewrite

Question 4

Factorise:

216x³ + Solution 4

216x³ + 

We know that 

Let us rewrite

Question 5

Factorise:

16x⁴ + 54xSolution 5

16x ⁴ + 54x

= 2x (8x ³ + 27)

= 2x [(2x)³ + (3)³]

= 2x (2x + 3) [(2x)² – 2x3 + 3²]    

=2x(2x+3)(4x² -6x +9)
Question 6

Factorise:

7a³ + 56b³Solution 6

7a³ + 56b³

= 7(a³ + 8b³)

= 7 [(a)³ + (2b)³]

= 7 (a + 2b) [a² – a 2b + (2b)²]    

= 7 (a + 2b) (a² – 2ab + 4b²).Question 7

Factorise:

x⁵ + x²Solution 7

x⁵ + x²

= x²(x³ + 1)

= x² (x + 1) [(x)² – x 1 + (1)²]     

= x² (x + 1) (x² – x + 1).Question 8

Factorise:

a³ + 0.008Solution 8

a³ + 0.008

= (a)³ + (0.2)³

= (a + 0.2) [(a)² – a 0.2 + (0.2)²]     

= (a + 0.2) (a² – 0.2a + 0.04).Question 9

Factorise:

1 – 27x³Solution 9

1 – 27x³

= (1)³ – (3x)³

= (1 – 3x) [(1)² + 1 3x + (3x)²]       

= (1 – 3x) (1 + 3x + 9x²).Question 10

Factorise 

64a³ – 343Solution 10

64a³ – 343

= (4a)³ – (7)³

= (4a – 7)[(4a)² + (4a)(7) + (7)²]

= (4a – 7)(16a² + 28a + 49) Question 11

Factorise:

x³ – 512Solution 11

x³ – 512

= (x)³ – (8)³

= (x – 8) [(x)² + x 8 + (8)²]       

= (x – 8) (x² + 8x + 64).Question 12

Factorise:

a³ – 0.064Solution 12

a³ – 0.064

= (a)³ – (0.4)³

= (a- 0.4) [(a)² + a 0.4 + (0.4)²]     

= (a – 0.4) (a² + 0.4 a + 0.16).Question 13

Factorise:

Solution 13

We know that 

Let us rewrite

Question 14

Factorise 

Solution 14

Question 15

Factorise:

x – 8xy³Solution 15

x – 8xy³

= x (1 – 8y³)

= x [(1)³ – (2y)³]

= x (1 – 2y) [(1)² + 1 2y + (2y)²]    

= x (1 – 2y) (1 + 2y + 4y²).Question 16

Factorise:

32x⁴ – 500xSolution 16

32x⁴ – 500x

= 4x (8x³ – 125)

= 4x [(2x)³ – (5)³]

= 4x [(2x – 5) [(2x)² + 2x 5 + (5)²]      

= 4x (2x – 5) (4x² + 10x + 25).Question 17

Factorise:

3a⁷b – 81a⁴b⁴Solution 17

3a⁷b – 81a⁴b⁴

= 3a⁴b (a³ – 27b³)

= 3a⁴b [(a)³ – (3b)³]

= 3a⁴b (a – 3b) [(a)² + a 3b + (3b)²]      

= 3a⁴b (a – 3b) (a² + 3ab + 9b²).Question 18

Factorise 

x⁴y⁴ – xySolution 18

x⁴y⁴ – xy

= xy(x³y³ – 1)

= xy[(xy)³ – (1)³]

= xy{(xy – 1)[(xy)² + (xy)(1) + (1)²]}

= xy(xy – 1)(x²y² + xy + 1) Question 19

Factorise 

8x²y³ – x⁵Solution 19

8x²y³ – x⁵

= x² (8y³ – x³)

= x² [(2y)³ – x³]

= x² [(2y – x)[(2y)² + (2y)(x) + x²]

= x² (2y – x)(4y² + 2xy + x²) Question 20

Factorise 

1029 – 3x³Solution 20

1029 – 3x³

= 3(343 – x³)

= 3[(7)³ – x³]

= 3[(7 – x)(7² + 7x + x²)]

= 3(7 – x)(49 + 7x + x²) Question 21

Factorise:

x⁶ – 729Solution 21

x⁶ – 729

= (x²)³ – (9)³

= (x² – 9) [(x²)² + x² 9 + (9)²]      

= (x² – 9) (x⁴ + 9x² + 81)

= (x + 3) (x – 3) [(x² + 9)² – (3x)²]

= (x + 3) (x – 3) (x² + 3x + 9) (x² – 3x + 9).Question 22

Factorise 

x⁹ – y⁹Solution 22

x⁹ – y⁹

= (x³)³ – (y³)³ 

= [(x³ – y³)][(x³)² + x³y³ + (y³)²]

= [(x – y)(x² + xy + y²)(x⁶ + x³y³ + y⁶)  Question 23

Factorise:

(a + b)³ – (a – b)³Solution 23

We know that, 

Therefore,

(a + b)³ – (a – b)³

= [a + b – (a – b)] [ (a + b)² + (a + b) (a – b) + (a – b)²] 

= (a + b – a + b) [ a² + b² + 2ab + a² – b² + a² + b² – 2ab]

= 2b (3a² + b²).Question 24

Factorise:

8a³ – b³ – 4ax + 2bxSolution 24

8a³ – b³ – 4ax + 2bx

= 8a³ – b³ – 2x (2a – b)

= (2a)³ – (b)³ – 2x (2a – b)

= (2a – b) [(2a)² + 2a b + (b)²] – 2x (2a – b)    

= (2a – b) (4a² + 2ab + b²) – 2x (2a – b)

= (2a – b) (4a² + 2ab + b² – 2x).Question 25

Factorise:

a³ + 3a²b + 3ab² + b³ – 8Solution 25

Question 26

Factorise:

Solution 26

We know that 

Question 27

Factorise:

2a³ + 16b³ – 5a – 10bSolution 27

2a³ + 16b³ – 5a – 10b

= 2 (a³ + 8b³) – 5 (a + 2b)

= 2 [(a)³ + (2b)³] – 5 (a + 2b)

= 2 (a + 2b) [(a)² – a 2b + (2b)² ] – 5 (a + 2b)     

= (a+ 2b) [2(a² – 2ab + 4b²) – 5]Question 28

Factorise 

a⁶ + b⁶Solution 28

a⁶ + b⁶

= (a²)³ + (b²)³ 

= (a² + b²)[(a²)² – (a²b²) + (b²)²]

= (a² + b²)(a⁴ – a²b² + b⁴) Question 29

Factorise 

a¹² – b¹²Solution 29

a¹² – b¹²

= (a⁶)² – (b⁶)² 

= (a⁶ – b⁶)(a⁶ + b⁶)

= [(a³)² – (b³)²][(a²)³ + (b²)³]

= (a³ – b³)(a³ + b³)[(a² + b²)(a⁴ – a²b² + b⁴)] 

= (a – b)(a² + ab + b²)(a + b)(a² – ab + b²)(a² + b²)(a⁴ – a²b² + b⁴)

= (a – b)(a + b)(a² + b²)(a² + ab + b²)(a² – ab + b²)(a⁴ – a²b² + b⁴)Question 30

Factorise 

x⁶ – 7x³ – 8Solution 30

Given equation is x⁶ – 7x³ – 8.

Putting x³ = y, we get

y² – 7y – 8

= y² – 8y + y – 8

= y(y – 8) + 1(y – 8)

= (y – 8)(y + 1)

= (x³ – 8)(x³ + 1)

= (x³ – 2³)(x³ + 1³)

= (x – 2)(x² + 2x + 4)(x + 1)(x² – x + 1)

= (x – 2)(x + 1)(x² + 2x + 4)(x² – x + 1) Question 31

Factorise 

x³ – 3x² + 3x + 7Solution 31

x³ – 3x² + 3x + 7

= x³ – 3x² + 3x – 1 + 8

= (x³ – 3x² + 3x – 1) + 8

= (x – 1)³ + 2³

= (x – 1 + 2)[(x – 1)² – (x – 1)(2) + 2²]

= (x + 1)(x² – 2x + 1 – 2x + 2 + 4)

= (x + 1)(x² – 4x + 7) Question 32

Factorise 

(x + 1)³ + (x – 1)³Solution 32

(x + 1)³ + (x – 1)³

= (x + 1 + x – 1)[(x + 1)² – (x + 1)(x – 1) + (x – 1)²]

= 2x(x² + 2x + 1 – x² + 1 + x² – 2x + 1)

= 2x(x² + 3)  Question 33

Factorise 

(2a + 1)³ + (a – 1)³Solution 33

(2a + 1)³ + (a – 1)³

= (2a + 1 + a – 1)[(2a + 1)² – (2a + 1)(a – 1) + (a – 1)²]

= (3a)[4a² + 4a + 1 – 2a² + 2a – a + 1 + a² – 2a + 1]

= 3a(3a² + 3a + 3)

= 9a(a² + a + 1)  Question 34

Factorise 

8(x + y)³ – 27(x – y)³Solution 34

8(x + y)³ – 27(x – y)³

= [2³ (x + y)³] – [3³ (x – y)³]

= [2(x + y) – 3(x – y)]{[2(x + y)]² + 2(x + y)3(x – y) + [3(x – y)]²}

= (2x + 2y – 3x + 3y){[4(x² + y² + 2xy)] + 6(x² – y²) + [9(x² + y² – 2xy]}

= (-x + 5y){4x² + 4y² + 8xy + 6x² – 6y² + 9x² + 9y² – 18xy}

= (-x + 5y)(19x² + 7y² – 10xy) Question 35

Factorise 

(x + 2)³ + (x – 2)³Solution 35

(x + 2)³ + (x – 2)³

= [(x + 2) + (x – 2)][(x + 2)² – (x + 2)(x – 2) + (x – 2)²]

=(2x)(x² + 4x + 4 – x² + 4 + x² – 4x + 4)

= 2x(x² + 12) Question 36

Factorise 

(x + 2)³ – (x – 2)³Solution 36

(x + 2)³ – (x – 2)³

= [(x + 2) – (x – 2)][(x + 2)² + (x + 2)(x – 2) + (x – 2)²]

= 4(x² + 4x + 4 + x² – 4 + x² – 4x + 4)

= 4(3x² + 4) Question 37

Prove that

Solution 37

Question 38

Prove that

Solution 38

Exercise Ex. 3G

Question 1

Find the product:

(x + y – z) (x² + y² + z² – xy + yz + zx)Solution 1

(x + y – z) (x² + y² + z² – xy + yz + zx)

= [x + y + (-z)] [(x)² + (y)² + (-z)² – (x) (y) – (y) (-z) – (-z) (x)]

= x³ + y³ – z³ + 3xyz.Question 2

Find the product.

(x – y – z)(x² + y² + z² + xy – yz + xz)Solution 2

(x – y – z)(x² + y² + z² + xy – yz + xz)

= x³ + xy² + xz² + x²y  – xyz + x²z – x²y – y³ – yz² – xy² + y²z – xyz – x²z – y²z – z³ – xyz + yz² – xz²

= x³ – y³ – z³ – xyz – xyz – xyz

= x³ – y³ – z³ – 3xyz Question 3

Find the product:

(x – 2y + 3) (x² + 4y² + 2xy – 3x + 6y + 9)Solution 3

(x – 2y + 3) (x² + 4y² + 2xy – 3x + 6y + 9)

= [x + (-2y) + 3] [(x)² + (-2y)² + (3) – (x) (-2y) – (-2y) (3) – (3) (x)]

= (a + b + c) (a² + b² + c² – ab – bc – ca)

= a³ + b³ + c³ – 3abc

Where, x = a, (-2y) = b and 3 = c

(x – 2y + 3) (x² + 4y² + 2xy – 3x + 6y + 9)

= (x)³ + (-2y)³ + (3)² – 3 (x) (-2y) (3)

= x³ – 8y³ + 27 + 18xy.Question 4

Find the product.

(3x – 5y + 4)(9x² + 25y² + 15xy + 20y – 12x + 16)

*Modified the questionSolution 4

(3x – 5y + 4)(9x² + 25y² + 15xy + 20y – 12x + 16)

= (3x + (-5y) + 4)[(3x)² + (-5y)² + (4)² – (3x)(-5y) – (-5y)(4) – (3x)(4)]

= (3x)³ + (-5y)³ + (4)³ – 3(3x)(-5y)(4)

= 27x³ – 125y³ + 64 + 180xy Question 5

Factorise:

125a³ + b³ + 64c³ – 60abcSolution 5

125a³ + b³ + 64c³ – 60abc

= (5a)³ + (b)³ + (4c)³ – 3 (5a) (b) (4c)

= (5a + b + 4c) [(5a)² + b² + (4c)² – (5a) (b) – (b) (4c) – (5a) (4c)]

[ a³ + b³ + c³ – 3abc = (a+ b + c) (a² + b² + c² – ab – bc – ca)]

= (5a + b + 4c) (25a² + b² + 16c² – 5ab – 4bc – 20ac).Question 6

Factorise:

a³ + 8b³ + 64c³ – 24abcSolution 6

a³ + 8b³ + 64c³ – 24abc

= (a)³ + (2b)³ + (4c)³ – 3 a 2b 4c

= (a + 2b + 4c) [a² + 4b² + 16c² – 2ab – 8bc – 4ca).Question 7

Factorise:

1 + b³ + 8c³ – 6bcSolution 7

1 + b³ + 8c³ – 6bc

= 1 + (b)³ + (2c)³ – 3 (b) (2c)

= (1 + b + 2c) [1 + b² + (2c)² – b – b 2c – 2c]

= (1 + b + 2c) (1 + b² + 4c² – b – 2bc – 2c).Question 8

Factorise:

216 + 27b³ + 8c³ – 108bcSolution 8

216 + 27b³ + 8c³ – 108bc

= (6)³ + (3b)³ + (2c)² – 3 6 3b 2c

= (6 + 3b + 2c) [(6)² + (3b)² + (2c)² – 6 3b – 3b 2c – 2c 6]

= (6 + 3b + 2c) (36 + 9b² + 4c² – 18b – 6bc – 12c).Question 9

Factorise:

27a³ – b³ + 8c³ + 18abcSolution 9

27a³ – b³ + 8c³ + 18abc

= (3a)³ + (-b)³ + (2c)³ + 3(3a) (-b) (2c)

= [3a + (-b) + 2c] [(3a)² + (-b)² + (2c)² – 3a (-b) – (-b) (2c) – (2c) (3a)]

= (3a – b + 2c) (9a² + b² + 4c² + 3ab + 2bc – 6ca).Question 10

Factorise:

8a³ + 125b³ – 64c³ + 120abcSolution 10

8a³ + 125b³ – 64c³ + 120abc

= (2a)³ + (5b)³ + (-4c)³ – 3 (2a) (5b) (-4c)

= (2a + 5b – 4c) [(2a)² + (5b)² + (-4c)² – (2a) (5b) – (5b) (-4c) – (-4c) (2a)]

= (2a + 5b – 4c) (4a² + 25b² + 16c² – 10ab + 20bc + 8ca).Question 11

Factorise:

8 – 27b³ – 343c³ – 126bcSolution 11

8 – 27b³ – 343c³ – 126bc

= (2)³ + (-3b)³ + (-7c)³ – 3(2) (-3b) (-7c)

= (2 – 3b – 7c) [(2)² + (-3b)² + (-7c)² – (2) (-3b) – (-3b) (-7c) – (-7c) (2)]

= (2 – 3b – 7c) (4 + 9b² + 49c² + 6b – 21bc + 14c).Question 12

Factorise:

125 – 8x³ – 27y³ – 90xySolution 12

125 – 8x³ – 27y³ – 90xy

= (5)³ + (-2x)³ + (-3y)³ – 3 (5) (-2x) (-3y)

= (5 – 2x – 3y) [(5)² + (-2x)² + (-3y)² – (5) (-2x) – (-2x) (-3y) – (-3y) (5)]

= (5 – 2x – 3y) (25 + 4x² + 9y² + 10x – 6xy + 15y).Question 13

Factorise:

Solution 13

Question 14

Factorise:

27x³ – y³ – z³ – 9xyzSolution 14

27x³ – y³ – z³ – 9xyz

= (3x)³ – y³ – z³ – 3(3x)(y)(z)

= (3x)³ + (-y)³ + (-z)³ – 3(3x)(-y)(-z)

= [3x + (-y) + (-z)][(3x)² + (-y)² + (-z)² – (3x)(-y) – (-y)(-z) – (3x)(-z)]

= (3x – y – z)(9x² + y² + z² + 3xy – yz + 3zx) Question 15

Factorise:

Solution 15

Question 16

Factorise:

Solution 16

Question 17

Factorise:

(a – b)³ + (b – c)³ + (c – a)³Solution 17

Putting (a – b) = x, (b – c) = y and (c – a) = z, we get,

(a – b)³ + (b – c)³ + (c – a)³

= x³ + y³ + z³, where (x + y + z) = (a – b) + (b – c) + (c – a) = 0

= 3xyz[ (x + y + z)= 0 (x³ + y³ + z³) = 3xyz]

= 3(a – b) (b – c) (c – a).Question 18

Factorise:

(a – 3b)³ + (3b – c)³ + (c – a)³Solution 18

Given equation is (a – 3b)³ + (3b – c)³ + (c – a)³

Now,

(a – 3b) + (3b – c) + (c – a)

= a – a – 3b + 3b – c + c

= 0

We know that if x + y + z = 0, then x³ + y³ + z³ = 3xyz

Hence,

(a – 3b)³ + (3b – c)³ + (c – a)³

= 3(a – 3b)(3b – c)(c – a)Question 19

Factorise:

(3a – 2b)³ + (2b – 5c)³ + (5c – 3a)³Solution 19

We have:

(3a – 2b) + (2b – 5c) + (5c – 3a) = 0

So, (3a – 2b)³ + (2b – 5c)³ + (5c – 3a)³

= 3(3a – 2b) (2b – 5c) (5c – 3a).Question 20

Factorise:

(5a – 7b)³ + (7b – 9c)³ + (9c – 5a)³Solution 20

Given equation is (5a – 7b)³ + (7b – 9c)³ + (9c – 5a)³

Now,

(5a – 7b) + (7b – 9c) + (9c – 5a)

= 5a – 7b + 7b – 9c + 9c – 5a

= 0

We know that if x + y + z = 0, then x³ + y³ + z³ = 3xyz

Hence,

(5a – 7b)³ + (7b – 9c)³ + (9c – 5a)³

= 3(5a – 7b)(7b – 9c)(9c – 5a) Question 21

Factorise:

a³ (b – c)³ + b³ (c – a)³ + c³ (a – b)³Solution 21

a³ (b – c)³ + b³ (c – a)³ + c³ (a – b)³

= [a (b – c)]³ + [b (c – a)]³ + [c (a – b)]³

Now, since, a (b – c) + b (c -a) + c (a – b)

= ab – ac + bc – ba + ca – bc = 0

So, a³ (b – c)³ + b³ (c – a)³ + c³ (a – b)³

= 3a (b – c) b (c – a) c (a – b)

= 3abc (a – b) (b – c) (c – a).Question 22(i)

Evaluate

(-12)³ + 7³ + 5³Solution 22(i)

Given equation is (-12)³ + 7³ + 5³

Now,

-12 + 7 + 5 = -12 + 12 = 0

We know that if x + y + z = 0, then x³ + y³ + z³ = 3xyz

Hence,

(-12)³ + 7³ + 5³

= 3(-12)(7)(5)

= -1260 Question 22(ii)

Evaluate

(28)³ + (-15)³ + (-13)³Solution 22(ii)

Given equation is (28)³ + (-15)³ + (-13)³

Now,

28 + (-15) + (-13) = 28 – 28 = 0

We know that if x + y + z = 0, then x³ +  y³ + z³ = 3xyz

Hence,

(28)³ + (-15)³ + (-13)³

= 3(28)(-15)(-13)

= 16380 Question 23

Prove that (a + b + c)³ – a³ – b³ – c³ = 3(a + b)(b + c)(c + a)Solution 23

L.H.S. = (a + b + c)³ – a³ – b³ – c³

= [(a + b) + c]³ – a³ – b³ – c³

= (a + b)³ + c³ + 3(a + b)c × [(a + b) + c] – a³ – b³ – c³

= a³ + b³ + 3ab(a + b) + c³ + 3(a + b)c × [(a + b) + c] – a³ – b³ – c³

= 3ab(a + b) + 3(a + b)c × [(a + b) + c]

= 3(a + b)[ab + c(a + b) + c²]

= 3(a + b)[ab + ac + bc + c²]

= 3(a + b)[a(b + c) + c(b + c)]

= 3(a + b)[(b + c)(a + c)]

= 3(a + b)(b + c)(c + a)

= R.H.S.Question 24

If a, b, c are all nonzero and a + b + c = 0, prove that

.Solution 24

Question 25

If a + b + c = 9 and a² + b² + c² = 35, find the value of (a³ + b³ + c³ – 3abc).Solution 25

a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)

= (a + b + c)[(a² + b² + c²) – (ab + bc + ca)] ….(i)

Now,

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

⇒ (9)² = 35 + 2(ab + bc + ca)

⇒ 81 = 35 + 2(ab + bc + ca)

⇒ 2(ab + bc + ca) = 46

⇒ ab + bc + ca = 23

Substituting in (i), we get

a³ + b³ + c³ – 3abc = (9)[35 – 23] = 9 × 12 = 108 

Exercise Ex. 3B

Question 1

Factorise:

9x² – 16y²Solution 1

9x² – 16y²

= (3x)² – (4y)²

= (3x + 4y)(3x – 4y) Question 2

Factorise:

Solution 2

Question 3

Factorise:

81 – 16x²Solution 3

81 – 16x²

= (9)² – (4x)²

= (9 – 4x)(9 + 4x) Question 4

Factorise:

5 – 20x²Solution 4

5 – 20x²

= 5(1 – 4x²)

= 5[(1)² – (2x)²]

= 5[(1 – 2x)(1 + 2x)]

= 5(1 – 2x)(1 + 2x) Question 5

Factorise:

2x⁴ – 32Solution 5

2x⁴ – 32

= 2(x⁴ – 16)

= 2[(x²)² – (4)²]

= 2[(x² – 4)(x² + 4)]

= 2[(x² – 2²)(x² + 4)]

= 2[(x – 2)(x + 2)(x² + 4)]

= 2(x – 2)(x + 2)(x² + 4) Question 6

Factorise:

3a³b – 243ab³Solution 6

3a³b – 243ab³

= 3ab (a² – 81 b²)

= 3ab [(a)² – (9b)²]

= 3ab (a + 9b) (a – 9b)Question 7

Factorise:

3x³ – 48xSolution 7

3x³ – 48x

= 3x (x² – 16)

= 3x [(x)² – (4)²]

= 3x (x + 4) (x – 4)Question 8

Factorise:

27a² – 48b²Solution 8

27a² – 48b²

= 3 (9a² – 16b²)

= 3 [(3a)² – (4b)²]

= 3(3a + 4b) (3a – 4b)Question 9

Factorise:

x – 64x³Solution 9

x – 64x³

= x (1 – 64x²)

= x[(1)² – (8x)²]

= x (1 + 8x) (1 – 8x)Question 10

Factorise:

8ab² – 18a³Solution 10

8ab² – 18a³

= 2a (4b² – 9a²)

= 2a [(2b)² – (3a)²]

= 2a (2b + 3a) (2b – 3a)Question 11

Factorise:

150 – 6x²Solution 11

150 – 6x²

= 6 (25 – x²)

= 6 (5² – x²)

= 6 (5 + x) (5 – x)Question 12

Factorise:

2 – 50x²Solution 12

2 – 50x²

= 2 (1 – 25x²)

= 2 [(1)² – (5x)²]

= 2 (1 + 5x) (1 – 5x)Question 13

Factorise:

20x² – 45Solution 13

20x² – 45

= 5(4x² – 9)

= 5 [(2x)² – (3)²]

= 5 (2x + 3) (2x – 3)Question 14

Factorise:

(3a + 5b)² – 4c²Solution 14

(3a + 5b)² – 4c²

= (3a + 5b)² – (2c)²

= (3a + 5b – 2c)(3a + 5b + 2c) Question 15

Factorise:

a² – b² – a – bSolution 15

a² – b² – a – b

= a² – b² – (a + b)

= (a – b)(a + b) – (a + b)

= (a + b)(a – b – 1) Question 16

Factorise:

4a² – 9b² – 2a – 3bSolution 16

4a² – 9b² – 2a – 3b

= (2a)² – (3b)² – (2a + 3b)

= (2a – 3b)(2a + 3b) – (2a + 3b)

= (2a + 3b)(2a – 3b – 1) Question 17

Factorise:

a² – b² + 2bc – c²Solution 17

a² – b² + 2bc – c²

= a² – (b² – 2bc + c²)

= a² – (b – c)²

= [a – (b – c)][a + (b – c)]

= (a – b + c)(a + b – c) Question 18

Factorise:

4a² – 4b² + 4a + 1Solution 18

4a² – 4b² + 4a + 1

= (4a² + 4a + 1) – 4b²

= [(2a)² + 2 × 2a × 1 + (1)²] – (2b)²

= (2a + 1)² – (2b)²

= (2a + 1 – 2b)(2a + 1 + 2b)

= (2a – 2b + 1)(2a + 2b + 1) Question 19

Factorise:

a² + 2ab + b² – 9c²Solution 19

a² + 2ab + b² – 9c²

= (a + b)² – (3c)²

= (a + b + 3c) (a + b – 3c)Question 20

Factorise:

108a² – 3(b – c)²Solution 20

108a² – 3(b – c)²

= 3 [(36a² – (b -c)²]

= 3 [(6a)² – (b – c)²]

= 3 (6a + b – c) (6a – b + c)Question 21

Factorise:

(a + b)³ – a – bSolution 21

(a + b)³ – a – b

= (a + b)³ – (a + b)

= (a + b) [(a + b)² – 1²]

= (a + b) (a + b + 1) (a + b – 1)Question 22

Factorise:

x² + y² – z² – 2xySolution 22

x² + y² – z² – 2xy

= (x² + y² – 2xy) – z²

= (x – y)² – z²

= (x – y – z)(x – y + z) Question 23

Factorise:

x² + 2xy + y² – a² + 2ab – b²Solution 23

x² + 2xy + y² – a² + 2ab – b²

= (x² + 2xy + y²) – (a² – 2ab + b²)

= (x + y)² – (a – b)²

= [(x + y) – (a – b)][(x + y) + (a – b)]

= (x + y – a + b)(x + y + a – b) Question 24

Factorise:

25x² – 10x + 1 – 36y²Solution 24

25x² – 10x + 1 – 36y²

= (25x² – 10x + 1) – 36y²

= [(5x)² – 2(5x)(1) + (1)²] – (6y)²

= (5x – 1)² – (6y)²

= (5x – 1 – 6y)(5x – 1 + 6y) Question 25

Factorise:

a – b – a² + b²Solution 25

a – b – a² + b²

= (a – b) – (a² – b²)

= (a – b) – (a – b) (a + b)

= (a – b) (1 – a – b)Question 26

Factorise:

a² – 4ac + 4c² – b²Solution 26

a² – 4ac + 4c² – b²

= a² – 4ac + 4c² – b²

= a² – 2 a 2c + (2c)² – b²

= (a – 2c)² – b²

= (a – 2c + b) (a – 2c – b)Question 27

Factorise:

9 – a² + 2ab – b²Solution 27

9 – a² + 2ab – b²

= 9 – (a² – 2ab + b²)

= 3² – (a – b)²

= (3 + a – b) (3 – a + b)Question 28

Factorise:

x³ – 5x² – x + 5Solution 28

x³ – 5x² – x + 5

= x² (x – 5) – 1 (x – 5)

= (x – 5) (x² – 1)

= (x – 5) (x + 1) (x – 1)Question 29

Factorise:

1 + 2ab – (a² + b²)Solution 29

1 + 2ab – (a² + b²)

= 1 – (a² + b² – 2ab)

= (1)² – (a – b)²

= [1 – (a – b)][1 + (a – b)]

= (1 – a + b)(1 + a – b) Question 30

Factorise:

9a² + 6a + 1 – 36b²Solution 30

9a² + 6a + 1 – 36b²

= (9a² + 6a + 1) – 36b²

= [(3a)² + 2(3a)(1) + (1)²] – (6b)²

= (3a + 1)² – (6b)²

= (3a + 1 – 6b)(3a + 1 + 6b) Question 31

Factorise:

x² – y² + 6y – 9Solution 31

x² – y² + 6y – 9

= x² – (y² – 6y + 9)

= x² – (y² – 2 y 3 + 3²)

= x² – (y – 3)²

= [x + (y – 3)] [x – (y – 3)] 

= (x + y – 3) (x – y + 3)Question 32

Factorise:

4x² – 9y² – 2x – 3ySolution 32

4x² – 9y² – 2x – 3y

= (2x)² – (3y)² – (2x + 3y)

= (2x + 3y) (2x – 3y) – (2x + 3y) 

= (2x + 3y) (2x – 3y – 1)Question 33

Factorise:

9a² + 3a – 8b – 64b²Solution 33

9a² + 3a – 8b – 64b²

= 9a² – 64b² + 3a – 8b

= (3a)² – (8b)² + (3a – 8b)

= (3a + 8b) (3a – 8b) + (3a – 8b)

= (3a – 8b) (3a + 8b + 1)Question 34

Factorise:

Solution 34

Question 35

Factorise:

Solution 35

Question 36

Factorise:

Solution 36

Question 37

Factorise:

x⁸ – 1Solution 37

x⁸ – 1

= (x⁴)² – (1)²

= (x⁴ – 1)(x⁴ + 1)

= [(x²)² – (1)²)(x⁴ + 1)

= (x² – 1)(x² + 1)(x⁴ + 1)

= (x – 1)(x + 1)(x² + 1)[(x²)² + (1)² + 2x² – 2x²

= (x – 1)(x + 1)(x² + 1)[(x²)² + (1)² + 2x²) – 2x²

Question 38

Factorise:

16x⁴ – 1Solution 38

16x⁴ – 1

= (4x²)² – (1)²

= (4x² – 1)(4x² + 1)

= [(2x)² – (1)²](4x² + 1)

= (2x – 1)(2x + 1)(4x² + 1) Question 39

Factorise:

81x⁴ – y⁴Solution 39

81x⁴ – y⁴

= (9x²)² – (y²)²

= (9x² – y²)(9x² + y²)

= [(3x)² – y²](9x² + y²)

= (3x – y)(3x + y)(9x² + y²) Question 40

Factorise:

x⁴ – 625Solution 40

x⁴ – 625

= (x²)² – (25)²

= (x² + 25) (x² – 25)

= (x² + 25) (x² – 5²)

= (x² + 25) (x + 5) (x – 5)

Exercise Ex. 3A

Question 1

Factorise:

9x² + 12xySolution 1

9x² + 12xy = 3x (3x + 4y)Question 2

Factorise:

18x²y – 24xyzSolution 2

18x²y – 24xyz = 6xy (3x – 4z)Question 3

Factorise:

27a³b³ – 45a⁴b²Solution 3

27a³b³ – 45a⁴b² = 9a³b² (3b – 5a)Question 4

Factorise:

2a (x+ y) – 3b (x + y)Solution 4

2a (x + y) – 3b (x + y) = (x + y) (2a – 3b)Question 5

Factorise:

2x (p² + q²) + 4y (p² + q²)Solution 5

2x (p² + q²) + 4y (p² + q²)

= (2x + 4y) (p² + q²)

= 2(x+ 2y) (p² + q²)Question 6

Factorise:

x (a – 5) + y (5 – a)Solution 6

x (a – 5) + y (5 – a)

= x (a – 5) + y (-1) (a – 5)

= (x – y) (a – 5)Question 7

Factorise:

4 (a + b) – 6 (a + b)²Solution 7

4 (a + b) – 6 (a + b)²

= (a + b) [4 – 6 (a + b)]

= 2 (a + b) (2 – 3a – 3b)

= 2 (a + b) (2 – 3a – 3b)Question 8

Factorise:

8 (3a – 2b)² – 10 (3a – 2b)Solution 8

8 (3a – 2b)² – 10 (3a – 2b)

= (3a – 2b) [8(3a – 2b) – 10]

= (3a – 2b) 2[4 (3a – 2b) – 5]

= 2 (3a – 2b) (12 a – 8b – 5)Question 9

Factorise:

x (x + y)³ – 3x²y (x + y)Solution 9

x (x + y)³ – 3x²y (x + y)

= x (x + y) [(x + y)² – 3xy]

= x (x + y) (x² + y² + 2xy – 3xy)

= x (x + y) (x² + y² – xy)Question 10

Factorise:

x³+ 2x² + 5x + 10Solution 10

x³+ 2x² + 5x + 10

= x² (x + 2) + 5 (x + 2)

= (x² + 5) (x + 2)Question 11

Factorise:

x² + xy – 2xz – 2yzSolution 11

x² + xy – 2xz – 2yz

= x (x + y) – 2z (x + y)

= (x+ y) (x – 2z)Question 12

Factorise:

a³b – a²b + 5ab – 5bSolution 12

a³b – a²b + 5ab – 5b

= a²b (a – 1) + 5b (a – 1)

= (a – 1) (a²b + 5b)

= (a – 1) b (a² + 5)

= b (a – 1) (a² + 5)Question 13

Factorise:

8 – 4a – 2a³ + a⁴Solution 13

8 – 4a – 2a³ + a⁴

= 4(2 – a) – a³ (2 – a)

= (2 – a) (4 – a³)Question 14

Factorise:

x³ – 2x²y + 3xy² – 6y³Solution 14

x³ – 2x²y + 3xy² – 6y³

= x² (x – 2y) + 3y² (x – 2y)

= (x – 2y) (x² + 3y²)Question 15

Factorise:

px – 5q + pq – 5xSolution 15

px + pq – 5q – 5x

= p(x + q) – 5 (q + x)

= (x + q) (p – 5)Question 16

Factorise:

x² + y – xy – xSolution 16

x² – xy + y – x

= x (x – y) – 1 (x – y)

= (x – y) (x – 1)Question 17

Factorise:

(3a – 1)² – 6a + 2Solution 17

(3a – 1)² – 6a + 2

= (3a – 1)² – 2 (3a – 1)

= (3a – 1) [(3a – 1) – 2]

= (3a – 1) (3a – 3)

= 3(3a – 1) (a – 1)Question 18

Factorise:

(2x – 3)² – 8x + 12Solution 18

(2x – 3)² – 8x + 12

= (2x – 3)² – 4 (2x – 3)

= (2x – 3) (2x – 3 – 4)

= (2x – 3) (2x – 7)Question 19

Factorise:

a³ + a – 3a² – 3Solution 19

a³ + a – 3a² – 3

= a(a² + 1) – 3 (a² + 1)

= (a – 3) (a² + 1)Question 20

Factorise:

3ax – 6ay – 8by + 4bxSolution 20

3ax – 6ay – 8by + 4bx

= 3a (x – 2y) + 4b (x – 2y)

= (x – 2y) (3a + 4b)Question 21

Factorise:

abx² + a²x + b²x +abSolution 21

abx² + a²x + b²x +ab

= ax (bx + a) + b (bx + a)

= (bx + a) (ax + b)Question 22

Factorise:

x³ – x² + ax + x – a – 1Solution 22

x³ – x² + ax + x – a – 1

= x³ – x² + ax – a + x – 1

= x² (x – 1) + a (x – 1) + 1 (x – 1)

= (x – 1) (x² + a + 1)Question 23

Factorise:

2x + 4y – 8xy – 1Solution 23

2x + 4y – 8xy – 1

= 2x – 1 – 8xy + 4y

= (2x – 1) – 4y (2x – 1)

= (2x – 1) (1 – 4y)Question 24

Factorise:

ab (x² + y²) – xy (a² + b²)Solution 24

ab (x² + y²) – xy (a² + b²)

= abx² + aby² – a²xy – b²xy

= abx² – a²xy + aby² – b²xy

= ax (bx – ay) + by(ay – bx)

= (bx – ay) (ax – by)Question 25

Factorise:

a² + ab (b + 1) + b³Solution 25

a² + ab (b + 1) + b³

= a² + ab² + ab + b³

= a² + ab + ab² + b³

= a (a + b) + b² (a + b)

= (a + b) (a + b²)Question 26

Factorise:

a³ + ab (1 – 2a) – 2b²Solution 26

a³ + ab (1 – 2a) – 2b²

= a³ + ab – 2a²b – 2b²

= a (a² + b) – 2b (a² + b)

= (a² + b) (a – 2b)Question 27

Factorise:

2a² + bc – 2ab – acSolution 27

2a² + bc – 2ab – ac

= 2a² – 2ab – ac + bc

= 2a (a – b) – c (a – b)

= (a – b) (2a – c)Question 28

Factorise:

(ax + by)² + (bx – ay)²Solution 28

(ax + by)² + (bx – ay)²

= a²x² + b²y² + 2abxy + b²x² + a²y² – 2abxy

= a²x² + b²y² + b²x² + a²y²

= a²x² + b²x² + b²y² + a²y²

= x² (a² + b²) + y²(a² + b²)

= (a² + b²) (x² + y²)Question 29

Factorise:

a (a + b – c) – bcSolution 29

a (a + b – c) – bc

= a² + ab – ac – bc

= a(a + b) – c (a + b)

= (a – c) (a + b)Question 30

Factorise:

a(a – 2b – c) + 2bcSolution 30

a(a – 2b – c) + 2bc

= a² – 2ab – ac + 2bc

= a (a – 2b) – c (a – 2b)

= (a – 2b) (a – c)Question 31

Factorise:

a²x² + (ax² + 1)x + aSolution 31

a²x² + (ax² + 1)x + a

= a²x² + ax³ + x + a

= ax² (a + x) + 1 (x + a)

= (ax² + 1) (a + x)Question 32

Factorise:

ab (x² + 1) + x (a² + b²)Solution 32

ab (x² + 1) + x (a² + b²)

= abx² + ab + a²x + b²x

= abx² + a²x + ab + b²x

= ax (bx + a) + b (bx + a)

= (bx + a) (ax + b)Question 33

Factorise:

x² – (a + b) x+ abSolution 33

x² – (a + b) x+ ab

= x² – ax – bx + ab

= x (x – a) – b(x – a)

= (x – a) (x – b)Question 34

Factorise:

Solution 34

Exercise MCQ

Question 1

Which of the following expressions is a polynomial in one variable?

Solution 1

Question 2

Which of the following expressions is a polynomial?

Solution 2

Question 3

Which of the following is a polynomial?

Solution 3

Question 4

Which of the following is a polynomial?

Solution 4

Question 5

Which of the following is a polynomial?

(a) x^-2 + x^-1 + 3

(b) x + x^-1 + 2

(c) x^-1

(d)0Solution 5

Question 6

Which of the following is a quadratic polynomial?

(a) x + 4

(b) x³ + x

(c) x³ + 2x + 6

(d)x² + 5x + 4Solution 6

Question 7

Which of the following is a linear polynomial?

(a) x + x²

(b) x + 1

(c) 5x² – x + 3

(d)Solution 7

Question 8

Which of the following is a binomial?

(a) x² + x + 3

(b) x² + 4

(c) 2x²

(d)Solution 8

Question 9

(a) 

(b) 2

(c) 1

(d)0Solution 9

Question 10

Degree of the zero polynomial is

(a) 1

(b) 0

(c) not defined

(d)none of theseSolution 10

Question 11

Zero of the zero polynomial is

(a) 0

(b) 1

(c) every real number

(d)not definedSolution 11

Question 12

If p(x) = x + 4, then p(x) + p(-x) = ?

(a) 0

(b) 4

(c) 2x

(d)8Solution 12

Question 13

Solution 13

Question 14

If p(x) = 5x – 4x² + 3 then p(-1) = ?

(a) 2

(b) -2

(c) 6

(d) -6Solution 14

Correct option: (d)

P(x) = 5x – 4x² + 3

⇒ p(-1) = 5(-1) – 4(-1)² + 3 = -5 – 4 + 3 = -6Question 15

If (x⁵¹ + 51) is divided by (x + 1) then the remainder is

(a) 0

(b) 1

(c) 49

(d) 50Solution 15

Correct option: (d)

Let f(x) = x⁵¹ + 51

By the remainder theorem, when f(x) is divided by (x + 1), the remainder is f(-1).

Now, f(-1) = [(-1)ⁿ + 51] = -1 + 51 = 50Question 16

If (x + 1) is a factor of the polynomial (2x² + kx) then k = ?

(a)   4

(b)   -3

(c)  2

(d) -2Solution 16

Correct option: (c)

Let p(x) = 2x² + kx

Since (x + 1) is a factor of p(x),

P(–1) = 0

⇒ 2(–1)² + k(–1) = 0

⇒ 2 – k = 0

⇒ k = 2Question 17

When p(x) = x⁴ + 2x³ – 3x² + x – 1 is divided by (x – 2), the remainder is

(a) 0

(b) -1

(c) -15

(d)21Solution 17

Question 18

When p(x) = x³ – 3x² + 4x + 32 is divided by (x + 2), the remainder is

(a) 0

(b) 32

(c) 36

(d)4Solution 18

Question 19

When p(x) = 4x³ – 12x² + 11x – 5 is divided by (2x – 1), the remainder is

(a) 0

(b) -5

(c) -2

(d)2Solution 19

Question 20

When p(x) =x³-ax²+x is divided by (x-a), the remainder is

(a) 0

(b) a

(c) 2a

(d)3aSolution 20

Question 21

When p(x) = x³ + ax² + 2x + a is divided by (x + a), the remainder is

(a) 0

(b) a

(c) -a

(d)2aSolution 21

Question 22

(x + 1) is a factor of the polynomial

(a) x³ – 2x² + x + 2

(b) x³ + 2x² + x – 2

(c) x³ + 2x² – x – 2

(d)x³ + 2x² – x + 2Solution 22

Question 23

Zero of the polynomial p(x) = 2x + 5 is

(a) 

(b) 

(c) 

(d) Solution 23

Correct option: (b)

p(x) = 2x + 5

Now, p(x) = 0

⇒ 2x + 5 = 0

⇒ 2x = -5

Question 24

The zeroes of the polynomial p(x) = x² + x – 6 are

(a) 2, 3

(b) -2, 3

(c) 2, -3

(d)-2, -3Solution 24

Question 25

The zeroes of the polynomial p(x) = 2x² + 5x – 3 are

Solution 25

Question 26

The zeros of the polynomial p(x) = 2x² + 7x – 4 are

(a) 

(b) 

(c) 

(d) Solution 26

Correct option: (c)

p(x) = 2x² + 7x – 4

Now, p(x) = 0

⇒ 2x² + 7x – 4 = 0

⇒ 2x² + 8x – x – 4 = 0

⇒ 2x(x + 4) – 1(x + 4) = 0

⇒ (x + 4)(2x – 1) = 0

⇒ x + 4 = 0 and 2x – 1 = 0

⇒ x = -4 and x =  Question 27

If (x + 5) is a factor of p(x) = x³ – 20x + 5k, then k =?

(a) -5

(b) 5

(c) 3

(d)-3Solution 27

Question 28

If (x + 2) and (x – 1) are factors of (x³ + 10x² + mx + n), then

(a) m = 5, n = -3

(b) m = 7, n = -18

(c) m = 17, n = -8

(d)m = 23, n = -19Solution 28

Question 29

If (x¹⁰⁰ + 2x⁹⁹ + k) is divisible by (x + 1), then the value of k is

(a) 1

(b) 2

(c) -2

(d)-3Solution 29

Question 30

For what value of k is the polynomial p(x) = 2x³ – kx² + 3x + 10 exactly divisible by (x + 2)?

Solution 30

Question 31

The zeroes of the polynomial p(x) = x² – 3x are

(a) 0, 0

(b) 0, 3

(c) 0, -3

(d)3, -3Solution 31

Question 32

The zeros of the polynomial p(x) = 3x² – 1 are

(a) 

(b) 

(c) 

(d) Solution 32

Correct option: (d)

p(x) = 3x² – 1

Now, p(x) = 0

⇒ 3x² – 1 = 0

⇒ 3x² = 1

Exercise Ex. 2A

Question 1(v)

Which of the expressions are polynomials?

Solution 1(v)

It is a polynomial, Degree = 2.Question 1(vi)

Which of the expressions are polynomials?

Solution 1(vi)

It is not a polynomial.Question 1(vii)

Which of the expressions are polynomials?

1Solution 1(vii)

It is a polynomial, Degree = 0.Question 1(viii)

Which of the expressions are polynomials?

Solution 1(viii)

It is a polynomial, Degree = 0.Question 1(i)

Which of the following expressions are polynomials? In case of a polynomial, write its degree.

Solution 1(i)

The given expression is an expression having only non-negative integral powers of x. So, it is a polynomial.

The highest power of x is 5. So, it is a polynomial of degree 5. Question 1(ii)

Which of the following expressions are polynomials? In case of a polynomial, write its degree.

Solution 1(ii)

The given expression is an expression having only non-negative integral powers of y. So, it is a polynomial.

The highest power of y is 3. So, it is a polynomial of degree 3. Question 1(iii)

Which of the following expressions are polynomials? In case of a polynomial, write its degree.

Solution 1(iii)

The given expression is an expression having only non-negative integral powers of t. So, it is a polynomial.

The highest power of t is 2. So, it is a polynomial of degree 2. Question 1(iv)

Which of the following expressions are polynomials? In case of a polynomial, write its degree.

X¹⁰⁰ – 1Solution 1(iv)

X¹⁰⁰ – 1 

The given expression is an expression having only non-negative integral powers of x. So, it is a polynomial.

The highest power of x is 100. So, it is a polynomial of degree 100. Question 1(ix)

Which of the following expressions are polynomials? In case of a polynomial, write its degree.

Solution 1(ix)

The given expression can be written as  

It contains a term having negative integral power of x. So, it is not a polynomial.Question 1(x)

Which of the following expressions are polynomials? In case of a polynomial, write its degree.

Solution 1(x)

The given expression is an expression having only non-negative integral powers of x. So, it is a polynomial.

The highest power of x is 2. So, it is a polynomial of degree 2. Question 1(xi)

Which of the following expressions are polynomials? In case of a polynomial, write its degree.

Solution 1(xi)

The given expression can be written as 2x^-2.

It contains a term having negative integral power of x. So, it is not a polynomial. Question 1(xii)

Which of the following expressions are polynomials? In case of a polynomial, write its degree.

Solution 1(xii)

The given expression contains a term containing x^1/2, where ½ is not a non-negative integer.

So, it is not a polynomial. Question 1(xiii)

Which of the following expressions are polynomials? In case of a polynomial, write its degree.

Solution 1(xiii)

The given expression is an expression having only non-negative integral powers of x. So, it is a polynomial.

The highest power of x is 2. So, it is a polynomial of degree 2. Question 1(xiv)

Which of the following expressions are polynomials? In case of a polynomial, write its degree.

x⁴ – x^3/2 + x – 3Solution 1(xiv)

x⁴ – x^3/2 + x – 3 

The given expression contains a term containing x^3/2, where 3/2 is not a non-negative integer.

So, it is not a polynomial. Question 1(xv)

Which of the following expressions are polynomials? In case of a polynomial, write its degree.

Solution 1(xv)

The given expression can be written as 2x³ + 3x² + x^1/2 – 1. 

The given expression contains a term containing x^1/2, where ½ is not a non-negative integer.

So, it is not a polynomial. Question 2(i)

Identify constant, linear, quadratic, cubic and quartic polynomials from the following.

-7 + xSolution 2(i)

-7 + x

The degree of a given polynomial is 1.

Hence, it is a linear polynomial.Question 2(ii)

Identify constant, linear, quadratic, cubic and quartic polynomials from the following.

6ySolution 2(ii)

6y

The degree of a given polynomial is 1.

Hence, it is a linear polynomial. Question 2(iii)

Identify constant, linear, quadratic, cubic and quartic polynomials from the following.

-z³Solution 2(iii)

-z³

The degree of a given polynomial is 3.

Hence, it is a cubic polynomial. Question 2(iv)

Identify constant, linear, quadratic, cubic and quartic polynomials from the following.

1 – y – y³Solution 2(iv)

1 – y – y³

The degree of a given polynomial is 3.

Hence, it is a cubic polynomial. Question 2(v)

Identify constant, linear, quadratic, cubic and quartic polynomials from the following.

x – x³ + x⁴Solution 2(v)

x – x³ + x⁴

The degree of a given polynomial is 4.

Hence, it is a quartic polynomial. Question 2(vi)

Identify constant, linear, quadratic, cubic and quartic polynomials from the following.

1 + x + x²Solution 2(vi)

1 + x + x²

The degree of a given polynomial is 2.

Hence, it is a quadratic polynomial. Question 2(vii)

Identify constant, linear, quadratic, cubic and quartic polynomials from the following.

-6x²Solution 2(vii)

-6x²

The degree of a given polynomial is 2.

Hence, it is a quadratic polynomial. Question 2(viii)

Identify constant, linear, quadratic, cubic and quartic polynomials from the following.

-13Solution 2(viii)

-13

The given polynomial contains only one term namely constant.

Hence, it is a constant polynomial. Question 2(ix)

Identify constant, linear, quadratic, cubic and quartic polynomials from the following.

-pSolution 2(ix)

-p

The degree of a given polynomial is 1.

Hence, it is a linear polynomial. Question 3(i)

Write the coefficient of x³ in x + 3x² – 5x³ + x⁴Solution 3(i)

The coefficient of x³ in x + 3x² – 5x³ + x⁴ is -5. Question 3(ii)

Write the coefficient of x in .Solution 3(ii)

The coefficient of x in . Question 3(iii)

Write the coefficient of x² in 2x – 3 + x³.Solution 3(iii)

The given polynomial can be written as x³ + 0x² + 2x – 3.

Hence, the coefficient of x² in 2x – 3 + x³ is 0. Question 3(iv)

Write the coefficient of x in .Solution 3(iv)

The coefficient of x in . Question 3(v)

Write the constant term in .Solution 3(v)

The constant term in . Question 4(i)

Determine the degree of each of the following polynomials.

Solution 4(i)

Hence, the degree of a given polynomial is 2. Question 4(ii)

Determine the degree of each of the following polynomials.

y²(y – y³)Solution 4(ii)

y²(y – y³)

= y³ – y⁵

Hence, the degree of a given polynomial is 5. Question 4(iii)

Determine the degree of each of the following polynomials.

(3x – 2)(2x³ + 3x²)Solution 4(iii)

(3x – 2)(2x³ + 3x²)

= 6x⁴ + 9x³ – 4x³ – 6x²

= 6x⁴ + 5x³ – 6x²

Hence, the degree of a given polynomial is 4. Question 4(iv)

Determine the degree of each of the following polynomials.

Solution 4(iv)

The degree of a given polynomial is 1. Question 4(v)

Determine the degree of each of the following polynomials.

-8Solution 4(v)

-8

This is a constant polynomial.

The degree of a non-zero constant polynomial is zero. Question 4(vi)

Determine the degree of each of the following polynomials.

x^-2(x⁴ + x²)Solution 4(vi)

x^-2(x⁴ + x²)

= x^-2.x²(x² + 1)

= x⁰ (x² + 1)

= x² + 1

Hence, the degree of a given polynomial is 2. Question 5(i)

Give an example of a monomial of degree 5.Solution 5(i)

Example of a monomial of degree 5:

3x⁵ Question 5(ii)

Give an example of a binomial of degree 8.Solution 5(ii)

Example of a binomial of degree 8:

x – 6x⁸ Question 5(iii)

Give an example of a trinomial of degree 4.Solution 5(iii)

Example of a trinomial of degree 4:

7 + 2y + y⁴ Question 5(iv)

Give an example of a monomial of degree 0.Solution 5(iv)

Example of a monomial of degree 0:

7 Question 6(i)

Rewrite each of the following polynomials in standard form.

x – 2x² + 8 + 5x³Solution 6(i)

x – 2x² + 8 + 5x³ in standard form:

5x³ – 2x² + x + 8 Question 6(ii)

Rewrite each of the following polynomials in standard form.

Solution 6(ii)

Question 6(iii)

Rewrite each of the following polynomials in standard form.

6x³ + 2x – x⁵ – 3x²Solution 6(iii)

6x³ + 2x – x⁵ – 3x² in standard form:

-x⁵ + 6x³ – 3x² + 2x Question 6(iv)

Rewrite each of the following polynomials in standard form.

2 + t – 3t³ + t⁴ – t²Solution 6(iv)

2 + t – 3t³ + t⁴ – t² in standard form:

t⁴ – 3t³ – t² + t + 2 

Exercise Ex. 2B

Question 1

If p(x) = 5 – 4x + 2x², find

(i) p(0)

(ii) p(3)

(iii) p(-2)Solution 1

p(x) = 5 – 4x + 2x²

(i) p(0) = 5 – 4  0 + 2  0² = 5

(ii) p(3) = 5 – 4  3 + 2  3²

= 5 – 12 + 18

= 23 – 12 = 11

(iii) p(-2) = 5 – 4(-2) + 2 (-2)²

= 5 + 8 + 8 = 21Question 2

If p(y) = 4 + 3y – y² + 5y³, find

(i) p(0)

(ii) p(2)

(iii) p(-1)Solution 2

p(y) = 4 + 3y – y² + 5y³

(i) p(0) = 4 + 3  0 – 0² + 5  0³

= 4 + 0 – 0 + 0 = 4

(ii) p(2) = 4 + 3  2 – 2² + 5  2³

= 4 + 6 – 4 + 40

= 10 – 4 + 40 = 46

(iii) p(-1) = 4 + 3(-1) – (-1)² + 5(-1)³

= 4 – 3 – 1 – 5 = -5Question 3

If f(t) = 4t² – 3t + 6, find

(i) f(0)

(ii) f(4)

(iii) f(-5)Solution 3

f(t) = 4t² – 3t + 6

(i) f(0) = 4  0² – 3  0 + 6

= 0 – 0 + 6 = 6

(ii) f(4) = 4(4)² – 3  4 + 6

= 64 – 12 + 6 = 58

(iii) f(-5) = 4(-5)² – 3(-5) + 6

= 100 + 15 + 6 = 121Question 4

If p(x) = x³ – 3x² + 2x, find p(0), p(1), p(2). What do you conclude?Solution 4

p(x) = x³ – 3x² + 2x

Thus, we have

p(0) = 0³ – 3(0)² + 2(0) = 0

p(1) = 1³ – 3(1)² + 2(1) = 1 – 3 + 2 = 0

p(2) = 2³ – 3(2)² + 2(2) = 8 – 12 + 4 = 0

Hence, 0, 1 and 2 are the zeros of the polynomial p(x) = x³ – 3x² + 2x.  Question 5

If p(x) = x³ + x² – 9x – 9, find p(0), p(3), p(-3) and p(-1). What do you conclude about the zeros of p(x)? Is 0 a zero of p(x)?Solution 5

p(x) = x³ + x² – 9x – 9

Thus, we have 

p(0) = 0³ + 0² – 9(0) – 9 = -9 

p(3) = 3³ + 3² – 9(3) – 9 = 27 + 9 – 27 – 9 = 0

p(-3) = (-3)³ + (-3)² – 9(-3) – 9 = -27 + 9 + 27 – 9 = 0

p(-1) = (-1)³ + (-1)² – 9(-1) – 9 = -1 + 1 + 9 – 9 = 0

Hence, 0, 3 and -3 are the zeros of p(x).

Now, 0 is not a zero of p(x) since p(0) ≠ 0. Question 6(i)

Verify that:

4 is a zero of the polynomial p(x) = x – 4.Solution 6(i)

p(x) = x – 4

Then, p(4) = 4 – 4 = 0

 4 is a zero of the polynomial p(x).Question 6(ii)

Verify that:

-3 is a zero of the polynomial p(x) = x – 3.Solution 6(ii)

p(x) = x – 3

Then,p(-3) = -3 – 3 = -6

 -3 is not a zero of the polynomial p(x).Question 6(iii)

Verify that:

is a zero of the polynomial p(y) = 2y + 1.Solution 6(iii)

p(y) = 2y + 1

Then, 

is a zero of the polynomial p(y).Question 6(iv)

Verify that:

is a zero of thepolynomial p(x) = 2 – 5x.Solution 6(iv)

p(x) = 2 – 5x

Then, 

 is a zero of the polynomial p(x).Question 7(i)

Verify that:

1 and 2 are the zeros of the polynomial p(x) = (x – 1) (x – 2).Solution 7(i)

p(x) = (x – 1) (x – 2)

Then,p(1)= (1 – 1) (1 – 2) = 0 -1 = 0

 1 is a zero of the polynomial p(x).

Also, p(2) = (2 – 1)(2 – 2) = 1 0 = 0

 2 is a zero of the polynomial p(x).

 Hence,1 and 2 are the zeroes of the polynomial p(x).Question 7(ii)

Verify that:

2 and -3 are the zeros of the polynomial p(x) = x² + x – 6.Solution 7(ii)

p(x) = x² + x – 6

Then, p(2) = 2² + 2 – 6

= 4 + 2 – 6

= 6 – 6 = 0

 2 is a zero of the polynomial p(x).

 Also, p(-3) = (-3)² – 3 – 6

= 9 – 3 – 6 = 0

 -3 is a zero of the polynomial p(x).

 Hence, 2 and -3 are the zeroes of the polynomial p(x).Question 7(iii)

Verify that:

0 and 3 are the zeros of the polynomial p(x) = x² – 3x.Solution 7(iii)

p(x) = x² – 3x.

Then,p(0) = 0² – 3 0 = 0

p(3) = (3)²– 3 3 = 9 – 9 = 0

0 and 3 are the zeroes of the polynomial p(x).Question 8(i)

Find the zero of the polynomial:

p(x) = x – 5Solution 8(i)

p(x) = 0

x – 5 = 0

x = 5

_{5 is the zero of the polynomial p(x).}Question 8(ii)

Find the zero of the polynomial:

q(x) = x + 4Solution 8(ii)

q(x) = 0

 x + 4 = 0

x= -4

 _{-4 is the zero of the polynomial q(x).} Question 8(iv)

Find the zero of the polynomial:

f(x) = 3x + 1Solution 8(iv)

f(x) = 0

 3x + 1= 0

3x=-1

 x =

  x =is the zero of the polynomial f(x).Question 8(v)

Find the zero of the polynomial:

g(x) = 5 – 4xSolution 8(v)

g(x) = 0

 5 – 4x = 0

 -4x = -5

 x =

 x =  is the zero of the polynomial g(x).Question 8(vii)

Find the zero of the polynomial:

p(x) = ax, a  0Solution 8(vii)

p(x) = 0

 ax = 0

x = 0

0 is the zero of the polynomial p(x).Question 8(viii)

Find the zero of the polynomial:

q(x) = 4xSolution 8(viii)

q(x) = 0

4x = 0

 x = 0

0 is the zero of the polynomial q(x).Question 8(iii)

Find the zero of the polynomial:

r(x) = 2x + 5Solution 8(iii)

r(x) = 2x + 5

Now, r(x) = 0

⇒ 2x + 5 = 0

⇒ 2x = -5

Question 8(vi)

Find the zero of the polynomial:

h(x) = 6x – 2Solution 8(vi)

h(x) = 6x – 2

Now, h(x) = 0

⇒ 6x – 2 = 0

⇒ 6x = 2

Question 9

If 2 and 0 are the zeros of the polynomial f(x) = 2x³ – 5x² + ax + b then find the values of a and b.

HINT f(2) = 0 and f(0) = 0.Solution 9

f(x) = 2x³ – 5x² + ax + b

Now, 2 is a zero of f(x).

⇒ f(2) = 0

⇒ 2(2)³ – 5(2)² + a(2) + b = 0

⇒ 16 – 20 + 2a + b = 0

⇒ 2a + b – 4 = 0 ….(i)

Also, 0 is a zero of f(x).

⇒ f(0) = 0

⇒ 2(0)³ – 5(0)² + a(0) + b = 0

⇒ 0 – 0 + 0 + b = 0

⇒ b = 0

Substituting b = 0 in (i), we get

2a + 0 – 4 = 0

⇒ 2a = 4

⇒ a = 2

Thus, a = 2 and b = 0. 

Exercise Ex. 2D

Question 1

Using factor theorem, show that:

(x – 2) is a factor of (x³ – 8)Solution 1

f(x) = (x³ – 8)

By the Factor Theorem, (x – 2) will be a factor of f(x) if f(2) = 0.

Here, f(2) = (2)³ – 8

= 8 – 8 = 0

 (x – 2) is a factor of (x³ – 8).
Question 2

Using factor theorem, show that:

(x – 3) is a factor of (2x³ + 7x² – 24x – 45)Solution 2

f(x) = (2x³ + 7x² – 24x – 45)

By the Factor Theorem, (x – 3) will be a factor of f(x) if f(3) = 0.

Here, f(3) = 2  3³ + 7  3² – 24  3 – 45

= 54 + 63 – 72 – 45

= 117 – 117 = 0

 (x – 3) is a factor of (2x³ + 7x² – 24x – 45).Question 3

Using factor theorem, show that:

(x – 1) is a factor of (2x⁴ + 9x³ + 6x² – 11x – 6)Solution 3

f(x) = (2x⁴ + 9x³ + 6x² – 11x – 6)

By the Factor Theorem, (x – 1) will be a factor of f(x) if f(1) = 0.

Here, f(1) = 2  1⁴ + 9  1³ + 6  1² – 11  1 – 6

= 2 + 9 + 6 – 11 – 6

= 17 – 17 = 0

 (x – 1) is factor of (2x⁴ + 9x³ + 6x² – 11x – 6).Question 4

Using factor theorem, show that:

(x + 2) is a factor of (x⁴ – x² – 12)Solution 4

f(x) = (x⁴ – x² – 12)

By the Factor Theorem, (x + 2) will be a factor of f(x) if f(-2) = 0.

Here, f(-2) = (-2)⁴ – (-2)² – 12

= 16 – 4 – 12

= 16 – 16 = 0

 (x + 2) is a factor of (x⁴ – x² – 12).Question 5

p(x) = 69 + 11x – x² + x³, g(x) = x + 3Solution 5

By the factor theorem, g(x) = x + 3 will be a factor of p(x) if p(-3) = 0.

Now, p(x) = 69 + 11x – x² + x³

⇒ p(-3) = 69 + 11(-3) – (-3)² + (-3)³

= 69 – 33 – 9 – 27

= 0

Hence, g(x) = x + 3 is a factor of the given polynomial p(x). Question 6

Using factor theorem, show that:

(x + 5) is a factor of (2x³ + 9x² – 11x – 30)Solution 6

f(x) = 2x³ + 9x² – 11x – 30

By the Factor Theorem, (x + 5) will be a factor of f(x) if f(-5) = 0.

Here, f(-5) = 2(-5)³ + 9(-5)² – 11(-5) – 30

= -250 + 225 + 55 – 30

= -280 + 280 = 0

 (x + 5) is a factor of (2x³ + 9x² – 11x – 30).Question 7

Using factor theorem, show that:

(2x – 3) is a factor of (2x⁴ + x³ – 8x² – x + 6)Solution 7

f(x) = (2x⁴ + x³ – 8x² – x + 6)

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0.

Here, 2x – 3 = 0  x = 

 is a factor of .Question 8

p(x) = 3x³ + x² – 20x + 12, g(x) = 3x – 2Solution 8

By the factor theorem, g(x) = 3x – 2 will be a factor of p(x) if  = 0.

Now, p(x) = 3x³ + x² – 20x + 12

Hence, g(x) = 3x – 2 is a factor of the given polynomial p(x). Question 9

Using factor theorem, show that:

(x – ) is a factor of Solution 9

f(x) = 

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0. 

Here, 

= 14 – 8 – 6

= 14 – 14 = 0

Question 10

Using factor theorem, show that:

(x + ) is a factor of Solution 10

f(x) = 

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0.

Here, 

Question 11

Show that (p – 1) is a factor of (p¹⁰ – 1) and also of (p¹¹ – 1).Solution 11

Let q(p) = (p¹⁰ – 1) and f(p) = (p¹¹ – 1)

By the factor theorem, (p – 1) will be a factor of q(p) and f(p) if q(1) and f(1) = 0.

Now, q(p) = p¹⁰ – 1

⇒ q(1) = 1¹⁰ – 1 = 1 – 1 = 0

Hence, (p – 1) is a factor of p¹⁰ – 1.

And, f(p) = p¹¹ – 1

⇒ f(1) = 1¹¹ – 1 = 1 – 1 = 0

Hence, (p – 1) is also a factor of p¹¹ – 1. Question 12

Find the value of k for which (x – 1) is a factor of (2x³+ 9x² + x + k).Solution 12

f(x) = (2x³ + 9x² + x + k)

x – 1 = 0  x = 1

 f(1) = 2  1³ + 9  1² + 1 + k

= 2 + 9 + 1 + k

= 12 + k

Given that (x – 1) is a factor of f(x).

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0 and therefore f(1) = 0.

f(1) = 12 + k = 0

k = -12.
Question 13

Find the value of a for which (x – 4) is a factor of (2x³ – 3x² – 18x + a).Solution 13

f(x) = (2x³ – 3x² – 18x + a)

x – 4 = 0  x = 4

 f(4) = 2(4)³ – 3(4)² – 18  4 + a

= 128 – 48 – 72 + a

= 128 – 120 + a

= 8 + a

Given that (x – 4) is a factor of f(x).

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0 and therefore f(4) = 0.

 f(4) = 8 + a = 0

 a = -8
Question 14

Find the value of a for which (x + 1) is a factor of (ax³ + x² – 2x + 4a – 9).Solution 14

Let p(x) = ax³ + x² – 2x + 4a – 9

It is given that (x + 1) is a factor of p(x).

⇒ p(-1) = 0

⇒ a(-1)³ + (-1)² – 2(-1) + 4a – 9 = 0

⇒ -a + 1 + 2 + 4a – 9 = 0

⇒ 3a – 6 = 0

⇒ 3a = 6

⇒ a = 2 Question 15

Find the value of a for which (x + 2a) is a factor of (x⁵ – 4a²x³ + 2x + 2a + 3).Solution 15

Let p(x) = x⁵ – 4a²x³ + 2x + 2a + 3

It is given that (x + 2a) is a factor of p(x).

⇒ p(-2a) = 0

⇒ (-2a)⁵ – 4a²(-2a)³ + 2(-2a) + 2a + 3 = 0

⇒ -32a⁵ – 4a²(-8a³) – 4a + 2a + 3 = 0

⇒ -32a⁵ + 32a⁵ -2a + 3 = 0

⇒ 2a = 3

Question 16

Find the value of m for which (2x – 1) is a factor of (8x⁴ + 4x³ – 16x² + 10x + m).Solution 16

Let p(x) = 8x⁴ + 4x³ – 16x² + 10x + m

It is given that (2x – 1) is a factor of p(x).

Question 17

Find the value of a for which the polynomial (x⁴ – x³ – 11x² – x + a) is divisible by (x + 3).Solution 17

Let p(x) = x⁴ – x³ – 11x² – x + a

It is given that p(x) is divisible by (x + 3).

⇒ (x + 3) is a factor of p(x).

⇒ p(-3) = 0

⇒ (-3)⁴ – (-3)³ – 11(-3)² – (-3) + a = 0

⇒ 81 + 27 – 99 + 3 + a = 0

⇒ 12 + a = 0

⇒ a = -12 Question 18

Without actual division, show that (x³ – 3x² – 13x + 15) is exactly divisible by (x² + 2x – 3).Solution 18

Let f(x) = x³ – 3x² – 13x + 15

Now, x² + 2x – 3 = x² + 3x – x – 3

= x (x + 3) – 1 (x + 3)

= (x + 3) (x – 1)

Thus, f(x) will be exactly divisible by x² + 2x – 3 = (x + 3) (x – 1) if (x + 3) and (x – 1) are both factors of f(x), so by factor theorem, we should have f(-3) = 0 and f(1) = 0.

Now, f(-3) = (-3)³ – 3 (-3)² – 13 (-3) + 15

= -27 – 3  9 + 39 + 15

= -27 – 27 + 39 + 15

= -54 + 54 = 0

And, f(1) = 1³ – 3  1² – 13  1 + 15

= 1 – 3 – 13 + 15

= 16 – 16 = 0

 f(-3) = 0 and f(1) = 0

So, x² + 2x – 3 divides f(x) exactly.Question 19

If (x³ + ax² + bx + 6) has (x – 2) as a factor and leaves a remainder 3 when divided by (x – 3), find the values of a and b.Solution 19

Letf(x) = (x³ + ax² + bx + 6)

Now, by remainder theorem, f(x) when divided by (x – 3) will leave a remainder as f(3).

So, f(3) = 3³ + a 3² + b 3 + 6 = 3

27 + 9a + 3b + 6 = 3

9 a + 3b + 33 = 3

9a + 3b = 3 – 33

9a + 3b = -30

3a + b = -10(i)

Given that (x – 2) is a factor of f(x).

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0 and therefore f(2) = 0.

 f(2) =  2³ + a 2² + b 2 + 6 = 0

                       8 + 4a+ 2b + 6 = 0

                               4a + 2b = -14

                                   2a + b = -7(ii)

Subtracting (ii) from (i), we get,

a = -3

Substituting the value of a = -3 in (i), we get,

3(-3) + b = -10

-9 + b = -10

b = -10 + 9

b = -1

a = -3 and b = -1.Question 20

Find the values of a and b so that the polynomial (x³ – 10x² + ax + b) is exactly divisible by (x – 1) as well as (x – 2).Solution 20

Let f(x) = (x³ – 10x² + ax + b), then by factor theorem

(x – 1) and (x – 2) will be factors of f(x) if f(1) = 0 and f(2) = 0.

f(1) = 1³ – 10 1² + a 1 + b = 0

1 – 10 + a + b = 0

a + b = 9(i)

Andf(2) = 2³ – 10 2² + a 2 + b = 0

8 – 40 + 2a + b = 0

2a + b = 32(ii)

Subtracting (i) from (ii), we get

a = 23

Substituting the value of a = 23 in (i), we get

23 + b = 9

b = 9 – 23

b = -14

a = 23 and b = -14.Question 21

Find the values of a and b so that the polynomial (x⁴ + ax³ – 7x² – 8x + b) is exactly divisible by (x + 2) as well as (x + 3).Solution 21

Letf(x)= (x⁴ + ax³ – 7x² – 8x + b)

Now, x + 2 = 0 x = -2 and x + 3 = 0 x = -3

By factor theorem, (x + 2) and (x + 3) will be factors of f(x) if f(-2) = 0 and f(-3) = 0

f(-2) = (-2)⁴ + a (-2)³ – 7 (-2)² – 8 (-2) + b = 0

16 – 8a – 28 + 16 + b = 0

-8a + b = -4

8a – b = 4(i)

And, f(-3) = (-3)⁴ + a (-3)³ – 7 (-3)² – 8 (-3) + b = 0

81 – 27a – 63 + 24 + b = 0

-27a + b = -42

27a – b = 42(ii)

Subtracting (i) from (ii), we get,

19a = 38

So, a = 2

Substituting the value of a = 2 in (i), we get

8 2 – b = 4

16 – b = 4

-b = -16 + 4

-b = -12

b = 12

a = 2 and b = 12.Question 22

If both (x – 2) and  are factors of px² + 5x + r, prove that p = r.Solution 22

Let f(x) = px² + 5x + r

Now, (x – 2) is a factor of f(x).

⇒ f(2) = 0

⇒ p(2)² + 5(2) + r = 0

⇒ 4p + 10 + r = 0

⇒ 4p + r = -10

Also,  is a factor of f(x).

From (i) and (ii), we have

4p + r = p + 4r

⇒ 4p – p = 4r – r

⇒ 3p = 3r

⇒ p = rQuestion 23

Without actual division, prove that 2x⁴ – 5x³ + 2x² – x + 2 is divisible by x² – 3x + 2.Solution 23

Let f(x) = 2x⁴ – 5x³ + 2x² – x + 2

and g(x) = x² – 3x + 2

= x² – 2x – x + 2

= x(x – 2) – 1(x – 2)

= (x – 2)(x – 1)

Clearly, (x – 2) and (x – 1) are factors of g(x).

In order to prove that f(x) is exactly divisible by g(x), it is sufficient to prove that f(x) is exactly divisible by (x – 2) and (x – 1).

Thus, we will show that (x – 2) and (x – 1) are factors of f(x).

Now,

f(2) = 2(2)⁴ – 5(2)³ + 2(2)² – 2 + 2 = 32 – 40 + 8 = 0 and

f(1) = 2(1)⁴ – 5(1)³ + 2(1)² – 1 + 2 = 2 – 5 + 2 – 1 + 2 = 0

Therefore, (x – 2) and (x – 1) are factors of f(x).

⇒ g(x) = (x – 2)(x – 1) is a factor of f(x).

Hence, f(x) is exactly divisible by g(x). Question 24

What must be added to 2x⁴ – 5x³ + 2x² – x – 3 so that the result is exactly divisible by (x – 2)?Solution 24

Let the required number to be added be k.

Then, p(x) = 2x⁴ – 5x³ + 2x² – x – 3 + k and g(x) = x – 2

Thus, we have

p(2) = 0

⇒ 2(2)⁴ – 5(2)³ + 2(2)² – 2 – 3 + k = 0

⇒ 32 – 40 + 8 – 5 + k = 0

⇒ k – 5 = 0

⇒ k = 5

Hence, the required number to be added is 5. Question 25

What must be subtracted from (x⁴ + 2x³ – 2x² + 4x + 6) so that the result is exactly divisible by (x² + 2x – 3)?Solution 25

Let p(x) = x⁴ + 2x³ – 2x² + 4x + 6 and q(x) = x² + 2x – 3.

When p(x) is divided by q(x), the remainder is a linear expression in x.

So, let r(x) = ax + b be subtracted from p(x) so that p(x) – r(x) is divided by q(x).

Let f(x) = p(x) – r(x) = p(x) – (ax + b)

= (x⁴ + 2x³ – 2x² + 4x + 6) – (ax + b)

= x⁴ + 2x³ – 2x² + (4 – a)x + 6 – b

We have,

q(x) = x² + 2x – 3

= x² + 3x – x – 3

= x(x + 3) – 1(x + 3)

= (x + 3)(x – 1)

Clearly, (x + 3) and (x – 1) are factors of q(x).

Therefore, f(x) will be divisible by q(x) if (x + 3) and (x – 1) are factors of f(x).

i.e., f(-3) = 0 and f(1) = 0

Consider, f(-3) = 0

⇒ (-3)⁴ + 2(-3)³ – 2(-3)² + (4 – a)(-3) + 6 – b = 0

⇒ 81 – 54 – 18 – 12 + 3a + 6 – b = 0

⇒ 3 + 3a – b = 0

⇒ 3a – b = -3 ….(i)

And, f(1) = 0

⇒ (1)⁴ + 2(1)³ – 2(1)² + (4 – a)(1) + 6 – b = 0

⇒ 1 + 2 – 2 + 4 – a + 6 – b = 0

⇒ 11 – a – b = 0

⇒ -a – b = -11 ….(ii)

Subtracting (ii) from (i), we get

4a = 8

⇒ a = 2

Substituting a = 2 in (i), we get

3(2) – b = -3

⇒ 6 – b = -3

⇒ b = 9

Putting the values of a and b in r(x) = ax + b, we get

r(x) = 2x + 9

Hence, p(x) is divisible by q(x), if r(x) = 2x + 9 is subtracted from it. Question 26

Use factor theorem to prove that (x + a) is a factor of (xⁿ + aⁿ) for any odd positive integer n.Solution 26

Let f(x) = xⁿ + aⁿ

In order to prove that (x + a) is a factor of f(x) for any odd positive integer n, it is sufficient to show that f(-a) = 0.

Now,

f(-a) = (-a)ⁿ + aⁿ

= (-1)ⁿ aⁿ + aⁿ

= [(-1)ⁿ + 1] aⁿ

= [-1 + 1] aⁿ …[n is odd ⇒ (-1)ⁿ = -1]

= 0 × aⁿ

= 0

Hence, (x + a) is a factor of xⁿ + aⁿ for any odd positive integer n. 

Exercise Ex. 2C

Question 1

By actual division, find the quotient and the remainder when (x⁴ + 1) is divided by (x – 1).

Verify that remainder = f(1).Solution 1

Quotient = x³ + x² + x + 1

Remainder = 2

Verification:

f(x) = x⁴ + 1

Then, f(1) = 1⁴ + 1 = 1 + 1 = 2 = Remainder Question 2

Verify the division algorithm for the polynomials

p(x) = 2x⁴ – 6x³ + 2x² – x + 2 and g(x) = x + 2.Solution 2

Question 3

Using remainder theorem, find the remainder when:

(x³ – 6x² + 9x + 3) is divided by (x – 1)Solution 3

f(x) = x³ – 6x² + 9x + 3

Now, x – 1 = 0  x = 1

By the remainder theorem, we know that when f(x) is divided by (x – 1) the remainder is f(1).

Now, f(1) = 1³ – 6  1² + 9  1 + 3

= 1 – 6 + 9 + 3

= 13 – 6 = 7

 The required remainder is 7.Question 4

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where

p(x) = 2x³ – 7x² + 9x – 13, g(x) = x – 3.Solution 4

x – 3 = 0

⇒ x = 3

By the remainder theorem, we know that when p(x) = 2x³ – 7x² + 9x – 13 is divided by g(x) = x – 3, the remainder is g(3).

Now,

g(3) = 2(3)³ – 7(3)² + 9(3) – 13 = 54 – 63 + 27 – 13 = 5

Hence, the required remainder is 5.Question 5

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where

p(x) = 3x⁴ – 6x² – 8x – 2, g(x) = x – 2.Solution 5

x – 2 = 0

⇒ x = 2

By the remainder theorem, we know that when p(x) = 3x⁴ – 6x² – 8x – 2 is divided by g(x) = x – 2, the remainder is g(2).

Now,

g(2) = 3(2)⁴ – 6(2)² – 8(2) – 2 = 48 – 24 – 16 – 2 = 6

Hence, the required remainder is 6. Question 6

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where

p(x) = 2x³ – 9x² + x + 15, g(x) = 2x – 3.Solution 6

2x – 3 = 0

⇒ x =  

By the remainder theorem, we know that when p(x) = 2x³ – 9x² + x + 15 is divided by g(x) = 2x – 3, the remainder is g.

Now,

Hence, the required remainder is 3. Question 7

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where

p(x) = x³ – 2x² – 8x – 1, g(x) = x + 1.Solution 7

x + 1 = 0

⇒ x = -1

By the remainder theorem, we know that when p(x) = x³ – 2x² – 8x – 1 is divided by g(x) = x + 1, the remainder is g(-1).

Now,

g(-1) = (-1)³ – 2(-1)² – 8(-1) – 1 = -1 – 2 + 8 – 1 = 4

Hence, the required remainder is 4. Question 8

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where

p(x) = 2x³ + x²– 15x – 12, g(x) = x + 2.Solution 8

x + 2 = 0

⇒ x = -2

By the remainder theorem, we know that when p(x) = 2x³ + x² – 15x – 12 is divided by g(x) = x + 2, the remainder is g(-2).

Now,

g(-2) = 2(-2)³ + (-2)² – 15(-2) – 12 = -16 + 4 + 30 – 12 = 6

Hence, the required remainder is 6. Question 9

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where

p(x) = 6x³ + 13x² + 3, g(x) = 3x + 2.Solution 9

3x + 2 = 0

⇒ x =  

By the remainder theorem, we know that when p(x) = 6x³ + 13x² + 3 is divided by g(x) = 3x + 2, the remainder is g.

Now,

Hence, the required remainder is 7. Question 10

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where

p(x) = x³ – 6x² + 2x – 4, g(x) = .Solution 10

By the remainder theorem, we know that when p(x) = x³ – 6x² + 2x – 4 is divided by g(x) = , the remainder is g.

Now,

Hence, the required remainder is . Question 11

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where

p(x) = 2x³ + 3x² – 11x – 3, g(x) = .Solution 11

By the remainder theorem, we know that when p(x) = 2x³ + 3x² – 11x – 3 is divided by g(x) = , the remainder is g.

Now,

Hence, the required remainder is 3. Question 12

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where

p(x) = x³ – ax² + 6x – a, g(x) = x – a.Solution 12

x – a = 0

⇒ x = a

By the remainder theorem, we know that when p(x) = x³ – ax² + 6x – a is divided by g(x) = x – a, the remainder is g(a).

Now,

g(a) = (a)³ – a(a)² + 6(a) – a = a³– a³+ 6a – a = 5a

Hence, the required remainder is 5a. Question 13

The polynomial (2x³ + x² – ax + 2) and (2x³ – 3x² – 3x + a) when divided by (x – 2) leave the same remainder. Find the value of a.Solution 13

Let p(x) = 2x³ + x² – ax + 2 and q(x) = 2x³ – 3x² – 3x + a be the given polynomials.

The remainders when p(x) and q(x) are divided by (x – 2) are p(2) and q(2) respectively.

By the given condition, we have

p(2) = q(2)

⇒ 2(2)³ + (2)² – a(2) + 2 = 2(2)³ – 3(2)² – 3(2) + a

⇒ 16 + 4 – 2a + 2 = 16 – 12 – 6 + a

⇒ 22 – 2a = -2 + a

⇒ a + 2a = 22 + 2

⇒ 3a = 24

⇒ a = 8 Question 14

The polynomial f(x) = x⁴ – 2x³ + 3x² – ax + b when divided by (x – 1) and (x + 1) leaves the remainders 5 and 19 respectively. Find the values of a and b. Hence, find the remainder when f(x) is divided by (x – 2).Solution 14

Letf(x) = (x⁴ – 2x³ + 3x² – ax + b)

_{From the given information,}

f(1) = 1⁴ – 2(1)³ + 3(1)² – a 1 + b = 5

 1 – 2 + 3 – a + b = 5

 2 – a + b = 5(i)

And,

f(-1) = (-1)⁴ – 2(-1)³ + 3(-1)² – a(-1) + b = 19

 1 + 2 + 3 + a + b = 19

 6 + a + b = 19(ii)

Adding (i) and (ii), we get

8 + 2b = 24

2b= 24 – 8 = 16

b = 

Substituting the value of b = 8 in (i), we get

2 – a + 8 = 5

-a + 10 = 5

-a = -10 + 5

-a = -5

a = 5

a = 5 and b = 8

f(x) = x⁴ – 2x³ + 3x² – ax + b

= x⁴ – 2x³ + 3x² – 5x + 8

f(2) = (2)⁴ – 2(2)³ + 3(2)² – 5 2 + 8

= 16 – 16 + 12 – 10 + 8

= 20 – 10 = 10

The required remainder is 10.Question 15

If p(x) = x³ – 5x² + 4x – 3 and g(x) = x – 2, show that p(x) is not a multiple of g(x).Solution 15

The polynomial p(x) will be a multiple of g(x) if g(x) divides p(x) completely.

i.e. when p(x) is divided by g(x), it does not leave any remainder.

Now, x – 2 = 0 ⇒ x = 2

Also,

p(2) = (2)³ – 5(2)² + 4(2) – 3 = 8 – 20 + 8 – 3 = -7 ≠ 0

Thus, p(x) is not a multiple of g(x).Question 16

If p(x) = 2x³ – 11x² – 4x + 5 and g(x) = 2x + 1, show that g(x) is not a factor of p(x).Solution 16

The polynomial g(x) will be a factor of p(x) if g(x) divides p(x) completely.

i.e. when p(x) is divided by g(x), it does not leave any remainder.

Now, 2x + 1 = 0 ⇒ x = 

Also,

Thus, g(x) is not a factor of p(x).