In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 2 Polynomials Ex 2.4 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board 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:
MP Board 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
MP Board 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
MP Board 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.

MP Board 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.
MP Board 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

MP Board 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.
MP Board Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.4 5

MP Board 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 ± \(\sqrt{3}\)
MP Board 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.

MP Board 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
MP Board 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
MP Board 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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