## MP Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.3

Question 1.

Find the remainder when x^{3} + 3x^{3} + 3x + 1 is divided by

- x + 1
- x – \(\frac{1}{2}\)
- x
- x + π
- 5 + 2x.

Solution:

Let p(x) = x^{3} + 3x^{2} + 3x + 1

1. x + 1

x + 1 = 0

x = – 1

Remainder

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

= – 1 + 3 – 3 + 1 = 0.

2. x – \(\frac{1}{2}\) = 0

x =π

When p(x) is divided by ( x – \(\frac{1}{2}\)) the remainder is P(\(\frac{1}{2}\))

3. x = 0

When p(x) is divided by x, the remainder is p(0).

p(0) = (0) + 3(0)^{2} + 3(0) + 1

= 0 + 0 + 0 + 1 = 1

4. x + π = 0

x = – π

When p(x) is divided by (x + π), the remainder is p(- π) p(- K) = (- π)^{3} + 3 (- π)^{2} + 3 (- π) + 1

= – π^{3} + 3π^{2} – 3π + 1

5. 5 + 2x = 0 x = \(\frac{-5}{2}\)

When p(x) is divided by (5 + 2x), the remainder is p(\(\frac{-5}{2}\))

Question 2.

Find the remainder when x^{3} – ax^{2} + 6x – a is divided by x – a.

Solution:

Let p(x) = x^{2} – ax^{2} + 6x – 1

x – a = 0

x = a

Remainder = (a)^{3} – a(a)^{2} + 6(a) – a

= a^{2} – a^{2} + 6a – a

= 5a.

Question 3.

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

Solution:

p(x) = 3x^{3} + 7x

7 + 3X = 0

(7 + 3x) is not a factor of p(x).

Factorization of Polynomials:

Factor Theorem:

Let p{x) be any polynomial of degree equal to or greater than 1 and ‘a’ be any real number such that –

- If p(a) = 0, then (x – a) is a factor of p(x)
- If (x – a) is a factor then p(a) = 0

Proof:

Let p(x) be a polynomial of degree ≥ 1 and be a real number.

1. If p(a) = 0, then (x – a) is a factor of p(x)

Suppose q(x) be the quotient when p(x) is divided by (x – a). By remainder theorem, remainder = p(a).

P(x) = (x – a) x q(x) + p(a)

p(x) = (x – a) x q(x) (∴ p(a) = 0, given)

= (x – a) is a factor of p(x).

2. If (x – a) is a factor then p(a) = 0

=> p(x) when divided by (x – a) gives remainder = 0 …(i)

But by remainder theorem

When p(x) is divided by (x – a), remainder = p(a) …(ii)

From (i) and (ii), we get

p(a) = 0 Proved.