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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Updated on: 10-Oct-2022

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