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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?

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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.

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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.

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You can do the polynomial division between $ p(x) $ and $ x^2-2x+1 $ or $x-1$ twice.You should obtain $rest=0$

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Consider $p(x) \in \mathbb{R}[x] $, holds the following theorem:

Theorem Let $p$ be a polynomial in $\mathbb{k}[x]$, the following condition are equivalent

  • $(x-a)| GCD(p,p')$, where $p'$ is the formal derivative of $p$ (in our case is the same of classical derivative)

  • $(x-a)^2 | p$

Now the derivative of your polynomial is $$p'(x)=(4-2a)x^3+(4-2a)x$$ Evaluating $p$ and $p'$ in 1 you obtain $p(1)=p'(1)=0$, so by the previous theorem you obtain the assert.

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Well, we can observe that $-ax^2+2ax-a=-a(x-1)^2,$ so our polynomial can be written as $$x^4-4x+3-a(x-1)^2.$$ All that remains is to show that $(x-1)^2$ is a factor of $x^4-4x+3,$ which is fairly straightforward to accomplish via polynomial long division, or by two applications of synthetic division.

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