Our ncert solutions for Solutions 2. Polynomials - Exercise 2.3 | Class 9 Mathematics - Toppers Study is the best material for English Medium students cbse board and other state boards students.

# Solutions 2. Polynomials - Exercise 2.3 | Class 9 Mathematics - Toppers Study

Topper Study classes prepares ncert solutions on practical base problems and comes out with the best result that helps the students and teachers as well as tutors and so many ecademic coaching classes that they need in practical life. Our ncert solutions for Solutions 2. Polynomials - Exercise 2.3 | Class 9 Mathematics - Toppers Study is the best material for English Medium students cbse board and other state boards students.

## 2. Polynomials

### | Exercise 2.3 |

## Solutions 2. Polynomials - Exercise 2.3 | Class 9 Mathematics - Toppers Study

**Chapter 2. Polynomials **

**Exercise 2.3**

**Solution: **

(i)By remainder theorem, the required remainder is equal to p(x) = (-1)

P (-1) = x^{3}+ 3x^{2}+ 3x+1

= (-1)^{3 }+ 3 (-1)^{2 }+ 3(-1) + 1

= -1 + 3 – 3 + 1 = 0

Required remainder is p (-1) = 0

Required remainder is p (1/2) = 0

(iv) By remainder theorem, the required remainder is equal to p(x) = - π

P(x) = x^{3}+ 3x^{2}+ 3x+1

P(π) = (- π)^{3 }+ 3(-π)^{2 }+ 3(-π) + 1

= (-π)^{3 }+ 3π^{2 }+ 3π + 1

Required remainder is p (π) = 0

Required remainder is p (- 5/2) = 0

**Q.2.** **find the remainder when x ^{3}- ax^{2 }+ 6x - a is divided by x- a.**

**Solution:** Let P(x) = x^{3}- ax^{2}+ 6x- a . P(x) = a.

By remainder theorem

P (a) = (a)^{3}- a(a)^{2}+ 6(a) - a

=a^{3}-a^{3 }+ 6a - a

Remainder = 5a

**Q3. Check whether 7 + 3x is a factor of 3x ^{3} + 7x. **

**Solution: **

g(x) = 7 + 3x = 0

⇒ 3x = - 7

P(x) ≠ 0

Therefore, 7 + 3x is not a factor of P(x).

##### Other Pages of this Chapter: 2. Polynomials

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