NCERT Solutions
Class 9 Maths
Polynomials

Ex.2.3 Q.1
Find the remainder when x3 + 3x2 + 3x + 1 is divided by
(1) x + 1
(2) x –
(3) x
(4) x + π
(5) 5 + 2x
Let p(x) = x3 + 3x2 + 3x + 1
(1) put x + 1 = 0, we get
x = -1
Using remainder theorem, when p(x) = x3 + 3x2 + 3x + 1 is divided by x + 1, remainder is given by p (-1).
Now, p (-1) = (-1)3 + 3 × (-1)2 + 3 × (-1) + 1
= -1 + 3 – 3 + 1
= -4 + 4
= 0
Hence, the remainder is 0.
(2) put x – = 0, we get
x =
Using remainder theorem, when p(x) = x3 + 3x2 + 3x + 1 is divided by x – , remainder is given
by p ( ).
Now, p ( ) = (
)3 + 3 × (
)2 + 3 × (
) + 1
= + 3 ×
+
+ 1
= +
+
+ 1
= (1 × 1 + 3 × 2 + 3 × 4 + 1 × 8) ÷ 8
[LCM (8, 4, 2, 1) = 8]
= (1 + 6 + 12 + 8) ÷ 8
=
Hence, the remainder is .
(3) put x = 0
Using remainder theorem, when p(x) = x3 + 3x2 + 3x + 1 is divided by x, remainder is given by p (0).
Now, p (-1) = (0)3 + 3 × (0)2 + 3 × (0) + 1
= 0 + 0 + 0 + 1
= 1
Hence, the remainder is 1.
(4) put x + π = 0
Using remainder theorem, when p(x) = x3 + 3x2 + 3x + 1 is divided by x + π, remainder is given
by p(-π).
Now, p(-π) = (-π)3 + 3 × (-π)2 + 3 × (-π) + 1
= -π3 + 3 π3 - 3 π + 1
Hence, the remainder is -π3 + 3 π3 - 3 π + 1.
(5) put 5 + 2x = 0, we get
2x = -5
=> x = -
Using remainder theorem, when p(x) = x3 + 3x2 + 3x + 1 is divided by 5 + 2x,
remainder is given by p (- ).
Now, p (- ) = (-
)3 + 3 × (-
)2 + 3 × (-
) + 1
= - + 3 ×
-
+ 1
= - +
-
+ 1
= (-1 × 125 + 75 × 2 - 15 × 4 + 1 × 8) ÷ 8
[LCM (8, 4, 2, 1) = 8]
= (-125 + 150 - 60 + 8) ÷ 8
= -
Hence, the remainder is .