Chapter 2 Polynomials Ex 2.4

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 – 26x² + 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.