Gujarat Board GSEB Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.3

Question 1.

Find the remainder when x^{3} + 3x^{2} + 3x + 1 is divided by

1. x + 1

2. x – \(\frac {1}{2}\)

3. x

4. x + π

5. 5 + 2x

Solutiobn:

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

1. Let x + 1 = 0

x = – 1

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

= -1 + 3 – 3 + 1= 0

∴ Remainder = 0

2. Let x – \(\frac {1}{2}\) = 0

x = \(\frac {1}{2}\)

Then p (\(\frac {1}{2}\)) = (\(\frac {1}{2}\)) + 3(\(\frac {1}{2}\)) + 3\(\frac {1}{2}\) + 1

= \(\frac {1}{8}\) + 3 x \(\frac {1}{4}\) + 3 x \(\frac {1}{2}\) + 1

= \(\frac {1}{8}\) + \(\frac {3}{4}\) + \(\frac {3}{2}\) + 1

= \(\frac{1+6+3 \times 4+1 \times 8}{8}\)

= \(\frac{1+6+12+8}{8}\)

= \(\frac {27}{8}\)

Hence, remainder = \(\frac {27}{8}\)

3. Let x = 0

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

∴ Remainder = 1

4. Let x + π = 0

x = – π

Then p(- π) = (-π)3 + 3(-π)2 + 3π + 1

= -π^{3} + 3π^{2} + 3π + 1

∴ Remainder = -π^{3} + 3π^{2} + 3π + 1

5. Let 5 + 2x = 0

2x = -5

x = \(\frac {-5}{2}\)

Then p(\(\frac {-5}{2}\)) = (\(\frac {-5}{2}\))^{3} + 3 (\(\frac {-5}{2}\))^{2} + 3(\(\frac {-5}{2}\)) + 1

= \(\frac {-125}{8}\) + \(\frac {3 x 25}{4}\) – \(\frac {15}{2}\) + 1

= \(\frac {-125}{8}\) + \(\frac {75}{4}\) – \(\frac {15}{2}\) + 1

= \(\frac{- 125 + 150 – 60 + 8}{8}\) = \(\frac {-27}{8}\)

∴ Remainder = \(\frac {-27}{8}\)

Question 2.

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

Solution:

Let p(x) = x^{3} – ax^{2} + 6x – a

and x – a = 0

= x = a

∴ p(a) = a^{3} – a x a^{2} + 6a – a

= a^{3} – a^{3} + 6a – a

p(a) = 5a

Hence remainder = 5a

Question 3.

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

Solution:

7 + 3x will be a factor of polynomial 3x^{3} + 7x if we divide 3x^{3} + 7x by 7 + 3x and it leaves no remainder.

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

and 7 + 3x = 0

3x = – 7

x = \(\frac {-7}{3}\)

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

So p(\(\frac {-7}{3}\)) = 3(\(\frac {-7}{3}\))^{3} + 7(\(\frac {-7}{3}\))

= 3 x \(\frac {-343}{9}\) – \(\frac {49}{3}\) = \(\frac {-343 – 147}{9}\) = \(\frac {-490}{9}\)

∴ p(x) = \(\frac {-490}{9}\)

Hence p(x) = 0

∴ 7 + 3 x is not a factor of 3 x^{3} + 7x.