Let $f(x)$ be a polynomial. If $(x-a)$ is a factor of $f(x)$, prove that if $f(x)$ is divided by $(x-a)^2$, the remainder will be equal to $n(x-a)$ for some real value $n$.
I have tried setting $f(x) $ to equal $g(x)(x-a)^2+k(x-a)$, but all that really proves that an equation with this form will satisfy it. I have no idea where to go from here, and to say that any equation following this form will work seems really flawed. Any help is appreciated.
Best Answer
Note that $f(x) = (x-a)g(x)$ for some $g(x)$ with degree $1$ less than $f(x)$. Also note that since $\deg(f)\geq 2, g(x)= (x-a)h(x) + r$ for some real $r$ and $h(x)$ with degree $2$ less than $f(x)$. Substituting back, we obtain:
$f(x) = h(x)(x - a)^{2} + r(x - a)$
Thus, $f(x)$'s remainder when divided by $(x-a)^{2}$ is of the form $r(x-a)$. $\blacksquare$