NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3 are part of 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

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 2 |

Polynomials | |

Exercise |
Ex 2.3 |

Number of Questions Solved |
5 |

Category |
NCERT Solutions |

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

**Question 1.
**Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

**(i)**p(x) = x

^{3}– 3x

^{2}+ 5x – 3, g(x) = x

^{2}– 2

**(ii)**p(x) = x

^{4}– 3x

^{2}+ 4x + 5, g(x) = x

^{2}+ 1 – x

**(iii)**p(x) = x

^{4}– 5x + 6, g(x) = 2 – x

^{2 }

**Solution:**

**Question 2.
**Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial.

**(i)**t

^{2}– 3, 2t

^{4}+ 3t

^{3}– 2t

^{2}– 9t – 12

**(ii)**x

^{2}+ 3x + 1, 3x

^{4}+ 5x

^{3}– 7x

^{2}+ 2x + 2

**(iii)**x

^{2}+ 3x + 1, x

^{5}– 4x

^{3 }+ x

^{2}+ 3x + 1

**Solution:**

∴ Remainder is 0, therefore, t

^{2}– 3 is a factor of polynomial 2t

^{4}+ 3t

^{3}– 2t

^{2}-9t – 12.

∴ Remainder is 0, therefore, x

^{2}+ 3x + 1 is a factor of polynomial 3x

^{4}+ 5x

^{3}– 7x

^{2}+ 2x + 2.

∴ Remainder = 2 ≠ 0, therefore, x

^{3}– 3x + 1 is not a factor of polynomial x

^{5}– 4x

^{3}+ x

^{2}+ 3x + 1.

**Question 3.
** Obtain all other zeroes of 3x

^{4}+ 6x

^{3}– 2x

^{2}– 10x – 5, if two of its zeroes are and and –

**Solution:**

= x (3x

^{2}– 5).Since both and(3x

^{2}– 5)are the factors, therefore 3x

^{2}– 5 is a factor of the given polynomial.

Now, we divide the given polynomial by 3x

^{2}– 5.

Hence, the other zeroes of the given polynomial are -1 and –1.

**Question 4.
**On dividing x

^{3 }– 3x

^{2}+ x + 2bya polynomial g(x), the quotient and remainder were x – 2 and -2x + 4 respectively. Find g(x).

**Solution:**

x

^{3}– 3x

^{2}+ x + 2 = g(x) x (x – 2) + (-2x + 4) [By division algorithm]

**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) = 2X

^{2}+ 2x + 8,

q(x) = x

^{2}+ x + 4,

g(x) = 2 and r(x) = 0

**(ii)** p(x) = x^{3} + x^{2} + x + 1,

q(x) = x + 1,

g(x) = x^{2} – 1 and r(x) = 2x + 2

**(iii)** p(x) = x^{3} – x^{2} + 2x + 3,

g(x) = x^{2} + 2,

q(x) = x – 1 and r(x) = 5

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