The problem states:
When $P(x)=ax^3+bx^2+cx+d$ is divided by $(x-a)$ the remainder is $p$, and when divided by $(x+a)$ the remainder is $-p$. Prove that $a=\frac{pb}{bc-ad}$.
I went ahead and started doing long division:
$$
\require{enclose}
\begin{array}{rll}
ax^2 && \hbox{} \\[-3pt]
(x-a) \enclose{longdiv}{ax^3+bx^2+cx+d}\kern-.2ex \\[-3pt]
\underline{ax^3-a^2x^2\phantom{3333333}} && \hbox{} \\[-3pt]
\phantom{0} && \hbox{} \\[-3pt]
\underline{\phantom{0}\phantom{0}} && \hbox{} \\[-3pt]
\phantom{0} && \hbox{} \\[-3pt]
\underline{\phantom{0}} && \hbox{} \\[-3pt]
\phantom{00}
\end{array}
$$
I stopped there didnt go any further because I know it will be hectic. Am I overthinking this? Please help.
In other words, you actually know that $P(a)=p$ and that $P(-a)=-p$. This tells you $$a^4+ba^2+ca+d=p$$ $$-a^4+ba^2-ca+d=-p$$
Summing both equalities, we obtain $2ba^2+2d=0$.
Now if $b=0$ we are done.
Otherwise, taking the difference of the two equations, we get $2a^4+2ca=2p$. We then write $a^4=(a^2)a^2=-\frac{da^2}{b}$. We obtain $a(c-\frac{da}{b})=p$, and we deduce then $$a=\frac{pb}{bc-ad}$$ (provided that $bc-ad$ is not zero).