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Home Class 10th Solutions 10th Maths

NCERT Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.4

by Sudhir
December 4, 2021
in 10th Maths, Class 10th Solutions
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In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 2 Polynomials Ex 2.4. These solutions are based on new NCERT Syllabus.

NCERT Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.4

Question 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; \(\frac{1}{2}\), 1, -2
(ii) x3 – 4x2 + 5x – 2; 2, 1, 1
Solution:
NCERT Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.4 1
Again, p(1) = 2(1)3 + (1)2 – 5(1) + 2
= 2 + 1 – 5 + 2 = 0
⇒ 1 is a zero of p(x).
Also, p(-2) = 2(-2)3 + (-2)2 – 5(-2) + 2
= -16 + 4 + 10 + 2 = -16 +16 = 0
= -2 is a zero of p(x).
Now, p(x) = 2x3 + x2 – 5x + 2
∴ Comparing it with ax3 + bx2 + cx + d, we have a = 2, b = 1, c = -5, and d = 2
NCERT Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.4 2
Thus, the relationship between the coefficients and the zeroes of p(x) is verified.

(ii) Here, p(x) = x3 – 4x2 + 5x – 2
∴ p(2) = (2)3 – 4(2)2 + 5(2) – 2
= 8 – 16 + 10 – 2 = 18 – 18 = 0
⇒ 2 is a zero of p(x)
Again p(1) = (1)3 – 4(1)2 + 5(1) – 2
= 1 – 4 + 5 – 2 = 6 – 6 = 0
⇒ 1 is a zero of p(x)
Now, comparing p(x) = x3 – 4x2 + 5x – 2
with ax3 + bx2 + cx + d = 0, we have
a = 1, b = -4, c = 5 and d = -2
Also 2, 1 and 1 are the zeroes of p(x).
Let α = 2,
β = 1,
γ = 1
Now, sum of zeroes = α + β + γ
= 2 + 1 + 1 = 4 = -b/a
NCERT Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.4 3
Thus, the relationship between the zeroes and the coefficients of p(x) is verified.

NCERT Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.4

Question 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.
Solution:
Let the required cubic polynomial be ax3 + 6x2 + cx + d = 0 and its zeroes be α, β and γ.
NCERT Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.4 4
∴ The requied cubic polynomial is
1x3 + (-2)x2 + (-7)x + 14 = 0
= x3 – 2x2 – 7x + 14 = 0

NCERT Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.4

Question 3.
If the zeroes of the polynomial x3 – 3x2 + x + 1 are a -b, a, a + b, find a and b.
Solution:
We have p(x) = x3 – 3x2 + x + 1
Comparing it with Ax3 + Bx2 + Cx + D,
We have A = 1, B = -3, C = 1 and D = 1
∵ It is given (a – b), a and (a + b) are the zeroes of the polynomial.
NCERT Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.4 5

NCERT Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.4

Question 4.
If two zeroes of the polynomial
x4 – 6x3 – 26x2+ 138x – 35 are 2±\(\sqrt{3}\), find other zeroes.
Solution:
Here, p(x) = x4 – 6x3 – 26x3 + 138x – 35
∵ Two of the zeroes of p(x) are : 2 ± \(\sqrt{3}\)
NCERT Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.4 6
(x2 – 4x + 1)(x2 – 2x – 35) = p(x)
⇒ (x2 – 4x + 1) (x2 – 7x + 5x – 35) = p(x)
⇒ (x2 – 4x + 1) [x(x – 7) + 5(x – 7)] = p(x)
⇒ (x2 – 4x + 1)(x – 7)(x + 5) = p(x)
i.e., (x – 7) and (x + 5) are other factors of p(x).
∴ 7 and – 5 are other zeroes of the given polynomial.

NCERT Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.4

Question 5.
If the polynomial x4 – 6x3 + 16x2 – 25x + 10 is divided by another polynomial x2 – 2x + k, the remainder comes out to be x + a, find k and a.
Solution:
Applying the division algorithm to the polynomials x4 – 6x3 + 16x2 – 25x + 10 and x2 – 2x + k, we have
NCERT Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.4 7
∴ Remainder = (2k – 9)x – k(8 – k) + 10
But the remainder = x + a (Given)
Therefore, comparing them, we have
NCERT Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.4 8
and a = -k(8 – k) + 10
= -5(8 – 5) + 10
= -5(3) + 10 = -15 + 10 = -5
Thus k = 5 and a = -5

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