In This Post we are providing Chapter 2 Polynomial NCERT Solutions for Class 10 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Polynomial Class 10 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.
We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 10 Maths Polynomial NCERT Written Solutions & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.
NCERT Solutions for Class 10 Maths Chapter 2 Polynomial
Exercises 2.1
1. The graphs of y = p(x) are given in following figure, for some polynomials p(x). Find the number of zeroes of p(x), in each case.
(i) The number of zeroes is 0 as the graph does not cut the x-axis at any point.
(ii) The number of zeroes is 1 as the graph intersects the x-axis at only 1 point.
(iii) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.
(iv) The number of zeroes is 2 as the graph intersects the x-axis at 2 points.
(v) The number of zeroes is 4 as the graph intersects the x-axis at 4 points.
(vi) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.
1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x2 – 2x – 8
(ii) 4s2 – 4s + 1
(iii) 6x2 – 3 – 7x
(iv) 4u2 + 8u
Answer
(i) x2 – 2x – 8
= (x – 4) (x + 2)
The value of x2 – 2x – 8 is zero when x – 4 = 0 or x + 2 = 0, i.e., when x = 4 or x = -2
Therefore, the zeroes of x2 – 2x – 8 are 4 and -2.
(ii) 4s2 – 4s + 1
= (2s-1)2
The value of 4s2 – 4s + 1 is zero when 2s – 1 = 0, i.e., s = 1/2
Sum of zeroes = 1/2 + 1/2 = 1 = -(-4)/4 = -(Coefficient of s)/Coefficient of s2
Product of zeroes = 1/2 × 1/2 = 1/4 = Constant term/Coefficient of s2.
(iii) 6x2 – 3 – 7x
= 6x2 – 7x – 3
= (3x + 1) (2x – 3)
The value of 6x2 – 3 – 7x is zero when 3x + 1 = 0 or 2x – 3 = 0, i.e., x = -1/3 or x = 3/2
Sum of zeroes = -1/3 + 3/2 = 7/6 = -(-7)/6 = -(Coefficient of x)/Coefficient of x2
Product of zeroes = -1/3 × 3/2 = -1/2 = -3/6 = Constant term/Coefficient of x2.
(iv) 4u2 + 8u
= 4u2 + 8u + 0
= 4u(u + 2)
The value of 4u2 + 8u is zero when 4u = 0 or u + 2 = 0, i.e., u = 0 or u = – 2
Therefore, the zeroes of 4u2 + 8u are 0 and – 2.
Sum of zeroes = 0 + (-2) = -2 = -(8)/4 = -(Coefficient of u)/Coefficient of u2
Product of zeroes = 0 × (-2) = 0 = 0/4 = Constant term/Coefficient of u2.
(v) t2 – 15
= t2 – 0.t – 15
= (t – √15) (t + √15)
The value of t2 – 15 is zero when t – √15 = 0 or t + √15 = 0, i.e., when t = √15 or t = -√15
Therefore, the zeroes of t2 – 15 are √15 and -√15.Sum of zeroes = √15 + -√15 = 0 = -0/1 = -(Coefficient of t)/Coefficient of t2
Product of zeroes = (√15) (-√15) = -15 = -15/1 = Constant term/Coefficient of u2.
(vi) 3x2 – x – 4
= (3x – 4) (x + 1)
The value of 3x2 – x – 4 is zero when 3x – 4 = 0 and x + 1 = 0,i.e., when x = 4/3 or x = -1
Therefore, the zeroes of 3x2 – x – 4 are 4/3 and -1.
2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) 1/4 , -1
Answer
(i) 1/4 , -1
Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = 1/4 = –b/a
αß = -1 = -4/4 = c/a
If a = 4, then b = -1, c = -4
Therefore, the quadratic polynomial is 4x2 – x -4.
(ii) √2 , 1/3
Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = √2 = 3√2/3 = –b/a
αß = 1/3 = c/a
If a = 3, then b = -3√2, c = 1
Therefore, the quadratic polynomial is 3x2 -3√2x +1.
(iii) 0, √5
Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = 0 = 0/1 = –b/a
αß = √5 = √5/1 = c/a
If a = 1, then b = 0, c = √5
Therefore, the quadratic polynomial is x2 + √5.
(iv) 1, 1
Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = 1 = 1/1 = –b/a
αß = 1 = 1/1 = c/a
If a = 1, then b = -1, c = 1
Therefore, the quadratic polynomial is x2 – x +1.
(v) -1/4 ,1/4
Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = -1/4 = –b/a
αß = 1/4 = c/a
If a = 4, then b = 1, c = 1
Therefore, the quadratic polynomial is 4x2 + x +1.
(vi) 4,1
Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = 4 = 4/1 = –b/a
αß = 1 = 1/1 = c/a
If a = 1, then b = -4, c = 1
Therefore, the quadratic polynomial is x2 – 4x +1.
1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:
Answer
(i) p(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2
second polynomial by the first polynomial:
Answer
p(x) = 3x4 + 6x3 – 2x2 – 10x – 5
Since the two zeroes are √(5/3) and – √(5/3).
Hence, the zeroes of the given polynomial are √(5/3) and – √(5/3), – 1 and – 1.
4. On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and
-2x + 4, respectively. Find g(x).
Answer
Here in the given question,
⇒ x3 – 3x2 + 3x – 2 = (x – 2) × g(x)
⇒ g(x) = (x3 – 3x2 + 3x – 2)/(x – 2)
5.Give examples of polynomial p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0
Answer
Here, p(x) = 6x2 + 2x + 2
g(x) = 2
q(x) = 3x2 + x + 1
r(x) = 0
Degree of p(x) and q(x) is same i.e. 2.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
Or, 6x2 + 2x + 2 = 2x (3x2 + x + 1)
Hence, division algorithm is satisfied.
(ii) Let us assume the division of x3+ x by x2,
Here, p(x) = x3 + x
g(x) = x2
q(x) = x and r(x) = x
Clearly, the degree of q(x) and r(x) is the same i.e., 1.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
x3 + x = (x2 ) × x + x
x3 + x = x3 + x
Thus, the division algorithm is satisfied.
(iii) Let us assume the division of x3+ 1 by x2.
Here, p(x) = x3 + 1
g(x) = x2
q(x) = x and r(x) = 1
Clearly, the degree of r(x) is 0.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
x3 + 1 = (x2 ) × x + 1
x3 + 1 = x3 + 1
Thus, the division algorithm is satisfied.
Exercise 2.4 (Optional)
1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:
(i) 2x3 + x2 – 5x + 2; 1/2, 1, -2
(ii) x3 – 4x2 + 5x – 2; 2, 1, 1
Answer
(i) p(x) = 2x3 + x2 – 5x + 2
Now for zeroes, putting the given value in x.
p(1/2) = 2(1/2)3 + (1/2)2 – 5(1/2) + 2
= (2×1/8) + 1/4 – 5/2 + 2
= 1/4 + 1/4 – 5/2 + 2
= 1/2 – 5/2 + 2 = 0
p(1) = 2(1)3 + (1)2 – 5(1) + 2
= (2×1) + 1 – 5 + 2
= 2 + 1 – 5 + 2 = 0
p(-2) = 2(-2)3 + (-2)2 – 5(-2) + 2
= (2 × -8) + 4 + 10 + 2
= -16 + 16 = 0
Thus, 1/2, 1 and -2 are the zeroes of the given polynomial.
Comparing the given polynomial with ax3 + bx2 + cx + d, we get a=2, b=1, c=-5, d=2
Also, α=1/2, β=1 and γ=-2
Now,
-b/a = α+β+γ
⇒ 1/2 = 1/2 + 1 – 2
⇒ 1/2 = 1/2
c/a = αβ+βγ+γα
⇒ -5/2 = (1/2 × 1) + (1 × -2) + (-2 × 1/2)
⇒ -5/2 = 1/2 – 2 – 1
⇒ -5/2 = -5/2
-d/a = αβγ
⇒ -2/2 = (1/2 × 1 × -2)
⇒ -1 = 1
Thus, the relationship between zeroes and the coefficients are verified.
(ii) p(x) = x3 – 4x2 + 5x – 2
Now for zeroes, putting the given value in x.
p(2) = 23 – 4(2)2 + 5(2) – 2
= 8 – 16 + 10 – 2
= 0
p(1) = 13 – 4(1)2 + 5(1) – 2
= 1 – 4 + 5 – 2
= 0
p(1) = 13 – 4(1)2 + 5(1) – 2
= 1 – 4 + 5 – 2
= 0
Thus, 2, 1 and 1 are the zeroes of the given polynomial.
Comparing the given polynomial with ax3 + bx2 + cx + d, we get a=1, b=-4, c=5, d=-2
Also, α=2, β=1 and γ=1
Now,
-b/a = α+β+γ
⇒ 4/1 = 2 + 1 + 1
⇒ 4 = 4
c/a = αβ+βγ+γα
⇒ 5/1 = (2 × 1) + (1 × 1) + (1 × 2)
⇒ 5 = 2 + 1 + 2
⇒ 5 = 5
-d/a = αβγ
⇒ 2/1 = (2 × 1 × 1)
⇒ 2 = 2
Thus, the relationship between zeroes and the coefficients are verified.
2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively.
Answer
Let the polynomial be ax3 + bx2 + cx + d and the zeroes be α, β and γ
Then, α + β + γ = -(-2)/1 = 2 = -b/a
αβ + βγ + γα = -7 = -7/1 = c/a
αβγ = -14 = -14/1 = -d/a
∴ a = 1, b = -2, c = -7 and d = 14
So, one cubic polynomial which satisfy the given conditions will be x3 – 2x2 – 7x + 14
3. If the zeroes of the polynomial x3 – 3x2 + x + 1 are a–b, a, a+b, find a and b.
Answer
Since, (a – b), a, (a + b) are the zeroes of the polynomial x3 – 3x2 + x + 1.
Therefore, sum of the zeroes = (a – b) + a + (a + b) = -(-3)/1 = 3
⇒ 3a = 3 ⇒ a =1
∴ Sum of the products of is zeroes taken two at a time = a(a – b) + a(a + b) + (a + b) (a – b) =1/1 = 1
a2 – ab + a2 + ab + a2 – b2 = 1
⇒ 3a2 – b2 =1
Putting the value of a,
⇒ 3(1)2 – b2 = 1
⇒ 3 – b2 = 1
⇒ b2 = 2
⇒ b = ±√2
Hence, a = 1 and b = ±√2
4. If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2±√3, find other zeroes.
Answer
2+√3 and 2-√3 are two zeroes of the polynomial p(x) = x4 – 6x3 – 26x2 + 138x – 35.
Let x = 2±√3
So, x-2 = ±√3
On squaring, we get x2 – 4x + 4 = 3,
⇒ x2 – 4x + 1= 0
Now, dividing p(x) by x2 – 4x + 1
∴ p(x) = x4 – 6x3 – 26x2 + 138x – 35
= (x2 – 4x + 1) (x2 – 2x – 35)
= (x2 – 4x + 1) (x2 – 7x + 5x – 35)
= (x2 – 4x + 1) [x(x – 7) + 5 (x – 7)]
= (x2 – 4x + 1) (x + 5) (x – 7)
∴ (x + 5) and (x – 7) are other factors of p(x).
∴ – 5 and 7 are other zeroes of the given polynomial.
Table of Contents
Important Links
Polynomial – Quick Revision Notes
Polynomial – Most Important Questions
Polynomial – Important MCQs
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