NCERT Solutions for Class 10 Maths 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;  \(\frac { 1 }{ 4 }\), 1, -2
(ii) x³ – 4x² + 5x – 2; 2, 1, 1
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


(ii) Compearing the given polynomial with ax³ + bx² + cx + d, we get:
a = 1, b = -4, c = 5 and d = – 2.
∴ p (x) = x³ – 4x² + 5x – 2
⇒  p(2) = (2)³ – 4(2)² + 5 x 2 – 2
= 8 – 16+ 10 – 2 = 0
p(1) = (1)³ – 4(1)² + 5 x 1-2
= 1 – 4 + 1 – 2
= 6-6 = 0
Hence, 2, 1 and 1 are the zeroes of x³ – 4x² + 5x – 2.
Hence verified.
Now we take α = 2, β = 1 and γ = 1.
α + β + γ = 2 + 1 + 1 = \(\frac { 4 }{ 1 }\) = \(\frac { -b }{ a }\)
αβ + βγ + γα = 2 + 1 + 2 = \(\frac { 5 }{ 1 }\) = \(\frac { c }{ a }\)
αβγ = 2 x 1 x 1  = \(\frac { 2 }{ 1 }\) = \(\frac { -d }{ a }\).
Hence 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.
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:
Let α , β and  γ be the zeroes of polynomial x³ – 3x² + x + 1.
Then α =  a-b, β = a and γ = a + b.
∴ Sum of zeroes = α + β + γ
⇒   3 = (a – b) + a + (a + b)
⇒  (a – b) + a + (a + b) = 3
⇒  a-b + a + a + b = 3
⇒       3a = 3
⇒ a =  \(\frac { 3 }{ 3 }\) = 1 …(i)
Product of zeroes = αβγ
⇒ -1 = (a – b) a (a + b)
⇒ (a – b) a (a + b) = -1
⇒   (a² – b²)a = -1
⇒  a³ – ab² = -1   … (ii)
Putting the value of a from equation (i) in equation (ii), we get:
(1)³-(1)b² = -1
⇒ 1 – b² = -1
⇒ – b² = -1 – 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, 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.

Using equation (i), we get:
(-9 + 2k)x + 10-8 k + k² = x + a
On comparing the like coefficients, we have:
-9 + 2k = 1
⇒ 2k = 10
⇒ k = \(\frac { 10 }{ 2 }\) = 5  ….(ii)
and 10 -8k + k²– a   ….(iii)
Substituting the value of k = 5, we get:
10 – 8(5) + (5)² = a
⇒   10 – 40 + 25 = a
⇒  35 – 40 =   a
⇒   a =   -5
Hence, k = 5 and a = -5.

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