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 x³ + 3x² + 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) = 0³ + 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 + 3 x is not a factor of 3 x³ + 7x.