Find in each case, the remainder when
(i) $x^{4} - 3 x^{2} + 2 x + 1$ is divided by x - 1
(ii) $x^{3} + 3 x^{2} - 12 x + 4 i s d i v i d e d b y x - 2$
(iii) $x^{4} + 1 i s d i v i d e d b y x + 1.$

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Solution

(i) $x^{4} - 3 x^{2} + 2 x + 1$ is divided by x - 1
x-1=0
x=1 put in equation
=> $(1)^{4} - 3 (1)^{2} + 2 (1) + 1$
=>1-3+2+1
=>1

(ii) $x^{3} + 3 x^{2} - 12 x + 4$ is divided by $x - 2$
x-2=0
x=2 put in equation
=> $(2)^{3} + 3 (2)^{2} - 12 (2) + 4$
=>8+12-24+4
=>0

(iii) $x^{4} + 1$ is divided by $x + 1$ .
x+1=0
x=-1 put in equation
=> $(- 1)^{4} + 1$
=>1+1
=>2

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