# NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3 are part ofÂ NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3.

• 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 2.3

Ex 2.3 Class 10 Maths Question 1.
Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:
(i)Â p(x) = x3Â â€“ 3x2Â + 5x â€“ 3, g(x) = x2Â â€“ 2
(ii)Â p(x) = x4Â â€“ 3x2Â + 4x + 5, g(x) = x2Â + 1 â€“ x
(iii)Â p(x) = x4â€“ 5x + 6, g(x) = 2 â€“ x2
Solution:

Ex 2.3 Class 10 MathsÂ Question 2.
Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial.
(i)Â t2Â â€“ 3, 2t4Â + 3t3Â â€“ 2t2â€“ 9t â€“ 12
(ii)Â x2Â + 3x + 1, 3x4Â + 5x3Â â€“ 7x2Â + 2x + 2
(iii)Â x2Â + 3x + 1, x5Â â€“ 4x3Â + x2Â + 3x + 1
Solution:

âˆ´Â  Remainder is 0, therefore, t2Â â€“ 3 is a factor of polynomial 2t4Â + 3t3Â â€“ 2t2Â -9t â€“ 12.

âˆ´ Remainder is 0, therefore, x2Â + 3x + 1 is a factor of polynomial 3x4Â + 5x3Â â€“ 7x2Â + 2x + 2.

âˆ´ Remainder = 2 â‰  0, therefore, x3Â â€“ 3x + 1 is not a factor of polynomial x5Â â€“ 4x3Â + x2+ 3x + 1.

Ex 2.3 Class 10 MathsÂ Question 3.
Â Obtain all other zeroes of 3x4Â + 6x3Â â€“ 2x2Â â€“ 10x â€“ 5, if two of its zeroes areÂ  andÂ $\sqrt { \frac { 5 }{ 3 } }$Â and â€“$\sqrt { \frac { 5 }{ 3 } }$
Solution:

=Â $\frac { 1 }{ 3 }$Â x (3x2â€“ 5).Since bothÂ $\frac { 1 }{ 3 }$Â and(3x2â€“ 5)are the factors, therefore 3x2Â â€“ 5 is a factor of the given polynomial.
Now, we divide the given polynomial by 3x2Â â€“ 5.

Hence, the other zeroes of the given polynomial areÂ -1Â andÂ â€“1.

Ex 2.3 Class 10 MathsÂ Question 4.
On dividing x3Â â€“ 3x2Â + x + 2bya polynomial g(x), the quotient and remainder were x â€“ 2 and -2x + 4 respectively. Find g(x).
Solution:
x3Â â€“ 3x2Â + x + 2Â  = g(x) x (x â€“ 2) + (-2x + 4) [By division algorithm]

Ex 2.3 Class 10 Maths Question 5.
Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and:
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0
Solution:
(i) p(x) = 2X2 + 2x + 8,
q(x) = x2 + x + 4,
g(x) = 2 and r(x) = 0

(ii) p(x) = x3 + x2 + x + 1,
q(x) = x + 1,
g(x) = x2 â€“ 1 and r(x) = 2x + 2

(iii) p(x) = x3 â€“ x2 + 2x + 3,
g(x) = x2 + 2,
q(x) = x â€“ 1 and r(x) = 5

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