# GSEB Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.3

Question 1.
Find the remainder when x3 + 3x3 + 3x + 1 is divided by
1. x + 1
2. x – $$\frac {1}{2}$$
3. x
4. x + π
5. 5 + 2x
Solutiobn:
p(x) = x3 + 3x2 + 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) = o3 + 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
= -π3 + 3π2 + 3π + 1
∴ Remainder = -π3 + 3π2 + 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 x3 – ax3 + 6x – a is divided by x – a.
Solution:
Let p(x) = x3 – ax2 + 6x – a
and x – a = 0
= x = a
∴ p(a) = a3 – a x a2 + 6a – a
= a3 – a3 + 6a – a
p(a) = 5a
Hence remainder = 5a

Question 3.
Check whether 7 + 3x is a factor of 3x3 + 7x.
Solution:
7 + 3x will be a factor of polynomial 3x3 + 7x if we divide 3x3 + 7x by 7 + 3x and it leaves no
remainder.
Let p(x) = 3x3 + 7x
and 7 + 3x = 0
3x = – 7
x = $$\frac {-7}{3}$$
Now, p(x) = 3x3 + 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 + 3x is not a factor of3x3 + 7x. 