# NCERT Solutions for Class 10 Maths Polynomials Exercise 2.4

In this page we have NCERT Solutions for Class 10 Maths Polynomials for EXERCISE 2.4 . Hope you like them and do not forget to like , social_share and comment at the end of the page.
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 - 5+ 2; 1/2, 1, -2
(ii) x3 - 4x2 + 5x - 2; 2, 1, 1
(i) p(x) = 2x3 + x2 - 5+ 2
Now for verification of 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 + c+ d, we get a=2, b=1, c=-5, d=2
Also, k1=1/2, k2=1 and k3=-2

Now,
-b/a = k1 + k2 +k3
⇒ 1/2 = 1/2 + 1 - 2
⇒ 1/2 = 1/2

c/a = k1k2+k2k3+k1k3
⇒ -5/2 = (1/2 × 1) + (1 × -2) + (-2 × 1/2)
⇒ -5/2 = 1/2 - 2 - 1
⇒ -5/2 = -5/2
-d/a = k1k2k3
⇒ -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 verification of 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 + c+ d, we get a=1, b=-4, c=5, d=-2
Also, k1=2, k2=1 and k3=1

Now,
-b/a = k1 + k2 +k3
⇒ 4/1 = 2 + 1 + 1
⇒ 4 = 4

c/a = k1k2+k2k3+k1k3
⇒ 5/1 = (2 × 1) + (1 × 1) + (1 × 2)
⇒ 5 = 2 + 1 + 2
⇒ 5 = 5
-d/a = k1k2k3
⇒ 2/1 = (2 × 1 × 1)
⇒ 2 = 2
Thus, the relationship between zeroes and the coefficients are verified.

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.
Let the polynomial be ax3 + bx+ cx + d and the zeroes be k1, k2 and k3

Then, k1 + k2 +k3 = -(-2)/1 = 2 = -b/a
k1k2+k2k3+k1k3= -7 = -7/1 = c/a
k1k2k3 = -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

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

Question 4
If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2±√3, find other zeroes.
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)
So (x + 5) and (x - 7) are other factors of p(x).
Therefore
- 5 and 7 are other zeroes of the given polynomial.

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.