Find all real numbers $a, b$ such that $(X-1)^{2}$ divides $a X^{4}+b X^{2}+1$.
The statement asks to find all real numbers $a, b$ such that the polynomial $P(x)=a x^{4}+b x^{2}+1$ has 1 as a double root, or equivalently, such that $P(1)=P^{\prime}(1)=0$, which can be rewritten as

$$
\begin{cases}a+b+1 & =0 \\ 4 a+2 b & =0\end{cases}
$$

We easily solve this system to obtain $(a, b)=(1,-2)$. We can verify by hand if we want (but it is not necessary since we reasoned by equivalences), that indeed,

$$
X^{4}-2 X^{2}+1=(X-1)^{2}(X+1)^{2}
$$