How to prove that $(x-1)^2$ is a factor of $x^4 - ax^2 + (2a-4)x + (3-a)$ for $a\in\mathbb R$?

Let $a \in R$. Verify that $(x − 1)^2$ is a factor of $$p(x) = x^4 − ax^2 + (2a − 4)x + (3 − a)$$

How can I solve this question?

Solution 1:

Hint apply the double root test (proof below)

$$\rm\begin{eqnarray} &&\rm\!\! (x\!-\!c)^2 |\ p(x)\!\!\!\!\!\!\!\\ \iff\ &&\rm x\!-\!c\ \ |\ \ p(x)\ &\rm and\ \ &\rm x\!-\!c\ \bigg|\ \dfrac{p(x)}{x\!-\!c}\\ \\ \iff\ &&\rm \color{#0a0}{p(c)} = 0 &\rm and&\rm x\!-\!c\ \bigg|\ \dfrac{p(x)-\color{#0a0}{p(c)}}{x\!-\!c}\ \ \left[\!\iff \color{#C00}{\dfrac{p(x)-p(c)}{x\!-\!c}\Bigg|_{\large\:x\:=\:c}} \!=\: 0\ \right] \\ \\ \iff\ &&\rm p(c) = 0 &\rm and&\rm \color{#C00}{p'(c)} = 0\end{eqnarray}$$

Remark $\ $ The proof is purely algebraic if you interpret the above $\rm\color{#c00}{red}$ expression as the algebraic definition of a polynomial derivative.

Solution 2:

You verify $p(1)=0$ and $q(1) = 0$ where $q(x) := p(x) / (x-1)$ is obtained by polynomial long division.

Another option is to verify $p(1) = p'(1) = 0$ (so $p$ 'touches' the $x$-Axis at $1$). These can be shown to be equivalent criteria.