Problem source

Prove that if ${(x-a)^{2}}$ is a factor of the polynomial $\p(x)$, then $\p'(a)=0$.
Prove a corresponding result if $(x-a)^4$ is a factor of $\p(x).$
Given that the polynomial 
$$
x^6+4x^5-5x^4-40x^3-40x^2+32x+k
$$
has a factor of the form ${(x-a)}^4$, find $k$.

Solution source

First notice that $p(x) = (x-a)^2q(x)$ so $p'(x) = 2(x-a)q(x) + (x-a)^2q'(x) = (x-a)(2q(x)+(x-a)q'(x))$, in particular $p'(a) = 0$ so $x-a$ is a root of $p'(x)$.

If $(x-a)^4$ is a root of $p(x)$ then $p^{(3)}(a)= 0$. The proof is similar.

Differentiating $3$ times we obtain:

$6 \cdot 5 \cdot 4 x^3 + 4 \cdot 5 \cdot 4 \cdot 3  x^2 - 5\cdot4 \cdot 3 \cdot 2 x-40 \cdot 3 \cdot 2 \cdot 1 = 5!(x^3+2x^2-x-2) = 5!(x+2)(x^2-1)$.

So our possible (repeated) roots are $x=-2,-1,1$.

We can check $p'(x) = 6x^5+20x^4-20x^3-120x^2-80x+32$, and see $p'(1) = 36 - 200 \neq 0$, $p'(-1) = -6+20+20-120+80+32 \neq 0$, therefore $a = -2$