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.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 2 |

Chapter Name |
Polynomials |

Exercise |
Ex 2.3 |

Number of Questions Solved |
4 |

Category |
NCERT Solutions |

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

**Ex 2.3 Class 10 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:**

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

^{3}-3x

^{2}+ 5x – 3 and g(x) = x

^{2}-2

dividing p(x) by g(x) ⇒

Quotient = x – 3, Remainder = 7x – 9

**(ii)** Here p(x) = x^{4}– 3x^{2} + 4x + 5 and g(x) = x^{2} + 1 -x

dividing p(x) by g(x) ⇒

Quotient = x^{2} + x – 3, Remainder = 8

**(iii)** Herep(x) = x^{4}– 5x + 6 and g(x) = 2-x^{2}

Rearranging g(x) = -x^{2} + 2

dividing p(x) by g(x) ⇒

Quotient = -x^{2} – 2

Remainder = -5x + 10.

**Ex 2.3 Class 10 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} + t^{3} – 2t^{2} – 9t – 12**

**(ii) x**

^{2}+ 3x + 1,3x^{4}+5x^{3}-7x^{2}+2x + 2**(iii) x**

^{3}-3x + 1, x^{5}– 4x^{3}+ x^{2}+ 3x + l**Solution:**

**(i)**First polynomial = t

^{2}– 3,

Second polynomial = 2t

^{4}+ 3t

^{3}– 2t

^{2}– 9t – 12

dividing second polynomial

by first polynomial ⇒

∵ Remainder is zero.

∴First polynomial is a factor of second polynomial.

**(ii)** First polynomial = x^{2} + 3x + 1

Second polynomial = 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2

dividing second polynomial

by first polynomial ⇒

∵ Remainder is zero.

∴ First polynomial is a factor of second polynomial.

**(iii)** First polynomial = x^{3} – 3x + 1

Second polynomial = x^{5} – 4x^{3} + x^{2} + 3x + 1.

∵ Remainder ≠ 0.

∴ First polynomial is not a factor of second polynomial.

**Ex 2.3 Solutions Class 10 Question 3.**

**Obtain all other zeroes of 3x ^{4} + 6x^{3} – 2x^{2} – 10x – 5, if two of its zeroes are \(\sqrt { \frac { 5 }{ 3 } }\) and –\(\sqrt { \frac { 5 }{ 3 } }\)**

**Solution:**

**NCERTSolutions Ex 2.3 Class 10 Question 4.**

**On dividing x ^{3} – 3x^{2} + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and -2x + 4, respectively. Find g(x).**

**Solution:**

p(x) = x

^{3}– 3 x 2 + x + 2 g(x) = ?

Quotient = x – 2; Remainder = -2x + 4

On dividing p(x) by g(x), we have

p(x) = g(x) x quotient + remainder

⇒ x

^{3}– 3x

^{2}+ x + 2 = g(x) (x – 2) + (-2x + 4)

⇒ x

^{3 }– 3x

^{2}+ x + 2 + 2 x- 4 = g(x) x (x-2)

⇒ x

^{3}– 3x

^{2}+ 3x – 2 = g(x) (x – 2)

**Exercise 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),g(x),q(x),r(x)

deg p(x) = deg q(x)

∴ both g(x) and r(x) are constant terms.

p(x) = 2x^{2}– 2x + 14; g(x) = 2

q(x) = x^{2} – x + 7; r(x) = 0

**(ii)** deg q(x) = deg r(x)

∴ this is possible when

deg of both q(x) and r(x) should be less than p(x) and g(x).

p(x) = x^{3}+ x^{2} + x + 1; g(x) = x^{2} – 1

q(x) = x + 1, r(x) = 2c + 2

**(iii)** deg r(x) is 0.

This is possible when product of q(x) and g(c) form a polynomial whose degree is equal to degree of p(x) and constant term.

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