Question:
Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
$p(x)=x^{3}-a x^{2}+6 x-a, g(x)=x-a$
Solution:
$p(x)=x^{3}-a x^{2}+6 x-a$
$g(x)=x-a$
By remainder theorem, when $p(x)$ is divided by $(x-a)$, then the remainder $=p(a)$.
Putting $x=a$ in $p(x)$, we get
$p(a)=a^{3}-a \times a^{2}+6 \times a-a=a^{3}-a^{3}+6 a-a=5 a$
$\therefore$ Remainder $=5 a$
Thus, the remainder when $p(x)$ is divided by $g(x)$ is $5 a$.
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