### 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:

Solution:
(i)
Comparing the given polynomial with ax3 + bx2 + cx + d, we get:
a = 2, b – 1, c = -5 and d = 2.
∴  p(x) = 2x3 + x2 – 5x + 2

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 α , β and  γ be the zeroes of the required polynomial.
Then α + β + γ = 2, αβ + βγ + γα = -7 and αβγ = -14.
∴ Cubic polynomial
= x3 – (α + β + γ)x2 + (αβ + βγ + γα)x – αβγ
= x3 – 2x2 – 1x + 14
Hence, the required cubic polynomial is x3 – 2x2 – 7x + 14.

Question 3.
If the zeroes of the polynomial x3 – 3x2 + x + 1 are a-b, a, a + b, find a and b.
Solution:

Question 4.
If two zeroes of the polynomial x4 – 6x3 – 26x+ 138x – 35 are 2 ± √3, finnd other zeroes.
Solution:
Since two zeroes are 2 + √3 and 2 – √3,
∴  [x-(2 + √3)] [x- (2 – √3)]
= (x-2- √3)(x-2 + √3)
= (x-2)2– (√3)2
x2 – 4x + 1 is a factor of the given polynomial.
Now, we divide the given polynomial by x2 – 4x + 1.

So, 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 – 7) (x + 5)
Hence, the other zeroes of the given polynomial are 7 and -5.

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:
We have
p(x) = x4 – 6x3 + 16x2 – 25x + 10
Remainder = x + a   … (i)
Now, we divide the given polynomial 6x3 + 16x2 – 25x + 10 by x2 – 2x + k.