Chapter 6 Factorisation of Polynomials Exercise Ex. 6.1
Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7
Solution 7
Question 8Give one example each of a binomial of degree 35, and of a monomial of degree 100.
Solution 8Degree of a polynomial is the highest power of variable in the polynomial.
Binomial has two terms in it. So binomial of degree 35 can be written as x35 + 7 . Monomial has only one term in it. So monomial of degree 100 can be written as 7x100. Concept Insight: Mono, bi and tri means one, two and three respectively. So, monomial is a polynomial having one term similarly for binomials and trinomials. Degree is the highest exponent of variable. The answer is not unique in such problems . Remember that the terms are always separated by +ve or -ve sign and not with .
Chapter 6 Factorisation of Polynomials Exercise Ex. 6.2
Question 1If f(x) = 2x3 – 13x2 + 17x + 12, find:
(i) f(2)
(ii) f(-3)
(iii) f(0)Solution 1(i)
f(x) = 2x3 – 13x2 + 17x + 12
f(2) = 2(2)3 – 13(2)2 + 17(2) + 12
= 16 – 52 + 34 + 12
= 10
(ii)
f(-3) = 2(-3)3 – 13(-3)2 + 17(-3) + 12
= -54 – 117 – 51 + 12
= – 210
(iii)
f(0) = 2(0)3 – 13(0)2 + 17(0) + 12
= 0 – 0 + 0 + 12
=12Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7Find rational roots of the polynomial f(x) = 2x3 + x2 – 7x – 6.Solution 7
Chapter 6 Factorisation of Polynomials Exercise Ex. 6.3
Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9
Question 10
If the polynomials ax3 + 3x2 – 13 and 2x3 -5x + a, when divided by (x-2) leave the same remainder, find the value of a.Solution 10
Question 11Find the remainder when x3 + 3x2 + 3x + 1 is divided by
Solution 11
Question 12
Solution 12
Chapter 6 Factorisation of Polynomials Exercise Ex. 6.4
Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9
Question 10
Solution 10
Question 11
Solution 11
Question 12
Solution 12
Question 13
Solution 13
Question 14
Solution 14
Question 15
Solution 15
Question 16
Solution 16
Question 17
Solution 17
Question 18
Solution 18
Question 19
Solution 19
Question 20
Solution 20
Question 21
Solution 21
Question 22
Solution 22
Question 23
Solution 23
Question 24
Solution 24
Question 25
What must be added to 3x3 + x2 – 22x + 9 so that the result is exactly divisible by 3x2 + 7x – 6?Solution 25
Chapter 6 Factorisation of Polynomials Exercise Ex. 6.5
Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Using factor theorem, factorize: x4 – 7x3 + 9x2 + 7x -10Solution 4
Question 5
Using factor theorem, factorize: 3x3 – x2 – 3x + 1Solution 5
Question 6
Using factor theorem, factorize each of the following polynomials:
x3 – 23x2 + 142x – 120Solution 6
Question 7
Using factor theorem, factorize: y3 – 7y + 6Solution 7
Question 8
Using factor theorem, factorize: x3 – 10x2 – 53x – 42Solution 8
Question 9
Using factor theorem, factorize: y3 – 2y2 – 29y – 42Solution 9
Question 10
Using factor theorem, factorize: 2y3 – 5y2 – 19y + 42Solution 10
Question 11
x3 + 13x2 + 32x + 20Solution 11
Let p(x) = x3 + 13x2 + 32x + 20
The factors of 20 are 1, 2, 4, 5 … …
By hit and trial method
p(- 1) = (- 1)3 + 13(- 1)2 + 32(- 1) + 20
= – 1 + 13 – 32 + 20
= 33 – 33 = 0
As p(-1) is zero, so x + 1 is a factor of this polynomial p(x).
Let us find the quotient while dividing x3 + 13x2 + 32x + 20 by (x + 1)
By long division
We know that
Dividend = Divisor Quotient + Remainder
x3 + 13x2 + 32x + 20 = (x + 1) (x2 + 12x + 20) + 0
= (x + 1) (x2 + 10x + 2x + 20)
= (x + 1) [x (x + 10) + 2 (x + 10)]
= (x + 1) (x + 10) (x + 2)
= (x + 1) (x + 2) (x + 10)Question 12
Factorise:
x3 – 3x2 – 9x – 5Solution 12
Let p(x) = x3 – 3x2 – 9x – 5
Factors of 5 are 1, 5.
By hit and trial method
p(- 1) = (- 1)3 – 3(- 1)2 – 9(- 1) – 5
= – 1 – 3 + 9 – 5 = 0
So x + 1 is a factor of this polynomial
Let us find the quotient while dividing x3 + 3x2 – 9x – 5 by x + 1
By long division
Now, Dividend = Divisor Quotient + Remainder
x3 – 3x2 – 9x – 5 = (x + 1) (x2 – 4 x – 5) + 0
= (x + 1) (x2 – 5 x + x – 5)
= (x + 1) [(x (x – 5) +1 (x – 5)]
= (x + 1) (x – 5) (x + 1)
= (x – 5) (x + 1) (x + 1) Question 13
Factorise:
2y3 + y2 – 2y – 1Solution 13
Let p(y) = 2y3 + y2 – 2y – 1 By hit and trial method
p(1) = 2 ( 1)3 + (1)2 – 2( 1) – 1
= 2 + 1 – 2 – 1= 0
So, y – 1 is a factor of this polynomial
By long division method,
p(y) = 2y3 + y2 – 2y – 1
= (y – 1) (2y2 +3y + 1)
= (y – 1) (2y2 +2y + y +1)
= (y – 1) [2y (y + 1) + 1 (y + 1)]
= (y – 1) (y + 1) (2y + 1)
Question 14
Using factor theorem, factorize: x3 – 2x2 – x + 2Solution 14
Question 15
Solution 15
Question 16
Using factor theorem, factroize : x4 – 2x3 – 7x2 + 8x + 12Solution 16
Question 17
Using factor theorem, factroize : x4 + 10x3 + 35x2 + 50x + 24Solution 17
Question 18
Using factor theorem, factorize : 2x4 – 7x3 – 13x2 + 63x – 45Solution 18
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