Using remainder theorem, find the remainder when: $f(x)=x^{2}+2ax+3a^{2},\ g( x)=x+a$.
Given: $f(x)=x^{2}+2ax+3a^{2},\ g( x)=x+a$.
To do: To find the remainder by using remainder theorem when $f( x)$ is divided by $g( x)$.
Solution:
As given, $f(x)=x^{2}+2ax+3a^{2}$ and $g( x)=x+a$.
Let $g( x)=x+a=0$
$\Rightarrow x=-a$, On substituting this value in $f( x)$.
$f( -a)=( -a)^2+2a( -a)+3a^2$
$\Rightarrow f( -a)=a^2-2a^2+3a^2$
$\Rightarrow f( -a)=2a^2$
Thus, the remainder is $2a^2$.
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