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