# NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4 are part ofÂ NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4.

• Polynomials Class 10 Ex 2.1
• Polynomials Class 10 Ex 2.2
• Polynomials Class 10 Ex 2.3
• Polynomials Class 10 Ex 2.4

## NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4

Ex 2.4 Class 10 Maths 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)Â 2x3Â + x2Â â€“ 5x + 2; Â $\frac { 1 }{ 4 }$, 1, -2
(ii)Â x3Â â€“ 4x2Â + 5x â€“ 2; 2, 1, 1
Solution:
(i)
Â Comparing the given polynomial with ax3Â + bx2Â + cx + d, we get:
a = 2, b â€“ 1, c = -5 and d = 2.
âˆ´Â Â p(x) = 2x3Â + x2Â â€“ 5x + 2

(ii)Â Compearing the given polynomial with ax3Â + bx2Â + cx + d, we get:
a = 1, b = -4, c = 5 and d = â€“ 2.
âˆ´ p (x) = x3Â â€“ 4x2Â + 5x â€“ 2
â‡’Â  p(2) = (2)3Â â€“ 4(2)2Â + 5 x 2 â€“ 2
= 8 â€“ 16+ 10 â€“ 2 = 0
p(1) = (1)3Â â€“ 4(1)2Â + 5 x 1-2
= 1 â€“ 4 + 1 â€“ 2
= 6-6 = 0
Hence, 2, 1 and 1 are the zeroes of x3Â â€“ 4x2Â + 5x â€“ 2.
Hence verified.
Now we take Î± = 2, Î² = 1 and Î³ = 1.
Î± + Î² + Î³ = 2 + 1 + 1 =Â $\frac { 4 }{ 1 }$Â =Â $\frac { -b }{ a }$
Î±Î² + Î²Î³ + Î³Î± = 2 + 1 + 2 =Â $\frac { 5 }{ 1 }$Â =Â $\frac { c }{ a }$
Î±Î²Î³ = 2 x 1 x 1Â  =Â $\frac { 2 }{ 1 }$Â =Â $\frac { -d }{ a }$.
Hence verified.

Ex 2.4 Class 10 MathsÂ 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 Î± , Î² andÂ  Î³ be the zeroes of the required polynomial.
Then Î± + Î² + Î³ = 2, Î±Î² + Î²Î³ + Î³Î± = -7 and Î±Î²Î³ = -14.
âˆ´ Cubic polynomial
= x3Â â€“ (Î± + Î² + Î³)x2Â + (Î±Î² + Î²Î³ + Î³Î±)x â€“ Î±Î²Î³
= x3Â â€“ 2x2Â â€“ 1x + 14
Hence, the required cubic polynomial is x3Â â€“ 2x2Â â€“ 7x + 14.

Ex 2.4 Class 10 MathsÂ Question 3.
If the zeroes of the polynomial x3Â â€“ 3x2Â + x + 1 are a-b, a, a + b, find a and b.
Solution:
Let Î± , Î² andÂ  Î³ be the zeroes of polynomial x3Â â€“ 3x2Â + x + 1.
Then Î± =Â  a-b, Î² = a and Î³ = a + b.
âˆ´ Sum of zeroes = Î± + Î² + Î³
â‡’Â Â  3 = (a â€“ b) + a + (a + b)
â‡’Â  (a â€“ b) + a + (a + b) = 3
â‡’Â  a-b + a + a + b = 3
â‡’ Â  Â  Â  3a = 3
â‡’ a =Â Â $\frac { 3 }{ 3 }$Â = 1 â€¦(i)
Product of zeroes = Î±Î²Î³
â‡’ -1 = (a â€“ b) a (a + b)
â‡’ (a â€“ b) a (a + b) = -1
â‡’Â Â  (a2Â â€“ b2)a = -1
â‡’ Â a3Â â€“ ab2Â = -1Â  Â â€¦ (ii)
Putting the value of a from equation (i) in equation (ii), we get:
(1)3-(1)b2Â = -1
â‡’ 1 â€“ b2Â = -1
â‡’ â€“ b2Â = -1 â€“ 1
â‡’Â Â b2Â = 2
â‡’ b = Â±âˆš2
Hence, a = 1 and b = Â±âˆš2.

Ex 2.4 Class 10 MathsÂ Question 4.
If two zeroes of the polynomial x4Â â€“ 6x3Â â€“ 26x2Â + 138x â€“ 35 are 2 Â± âˆš3, finnd other zeroes.
Solution:
Since two zeroes are 2 + âˆš3 and 2 â€“ âˆš3,
âˆ´Â  [x-(2 + âˆš3)] [x- (2 â€“ âˆš3)]
= (x-2- âˆš3)(x-2 + âˆš3)
= (x-2)2â€“ (âˆš3)2
x2Â â€“ 4x + 1 is a factor of the given polynomial.
Now, we divide the given polynomial by x2Â â€“ 4x + 1.

So, x4Â â€“ 6x3Â â€“ 26x2Â + 138x â€“ 35
= (x2Â â€“ 4x + 1) (x2Â â€“ 2x â€“ 35)
= (x2Â â€“ 4x + 1) (x2Â â€“ 7x + 5x â€“ 35)
= (x2-4x + 1) [x(x- 7) + 5 (x-7)]
= (x2Â â€“ 4x + 1) (x â€“ 7) (x + 5)
Hence, the other zeroes of the given polynomial are 7 and -5.

Ex 2.4 Class 10 Maths Question 5.
If the polynomial x4 â€“ 6x3 + 16x2 â€“ 25x + 10 is divided by another polynomial x2 â€“ 2x + k, the remainder comes out to be x + a, find k and a.
Solution:
We have
p(x) = x4 â€“ 6x3 + 16x2 â€“ 25x + 10
Remainder = x + a   â€¦ (i)
Now, we divide the given polynomial 6x3 + 16x2 â€“ 25x + 10 by x2 â€“ 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
â‡’ 2k = 10
â‡’ k = $\frac { 10 }{ 2 }$ = 5  â€¦.(ii)
and 10 -8k + k2â€“ a   â€¦.(iii)
Substituting the value of k = 5, we get:
10 â€“ 8(5) + (5)2 = a
â‡’   10 â€“ 40 + 25 = a
â‡’  35 â€“ 40 =   a
â‡’   a =   -5
Hence, k = 5 and a = -5.

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