Factor $x^5-5x^3+4x$

Question

I am trying to factor$x^5-5x^3+4x$ so that I can find the roots. I know from the answers section that the roots are where $x = 0, 1, -1, 2$ and $-2$.

I'm stuck, here's as far as I got:

$$ x^5-5x^3+4x = x(x^4-5x^2+4) $$

Let $u = x^2$ and just focus on the term on the right (drop the first $x$ for now):

$$x^4-5x^2+4 = u^2-5u+4x.$$

Master term is $a \times c = 1 \times 4 = 4$.

Seeking a pair of numbers that sum to the middle term $-5$ and whose product is $4$:

\begin{align} 1 \times -4 &= -4, &\text{sum} &= -3 \\ 4 \times -1 &= -4, &\text{sum} &= 3 \\ 2 \times -2 &= -4, &\text{sum} &= 0 \end{align}

???

I'm not sure how to proceed since I cannot find a pair of numbers that satisfy the condition.

Have I gone wrong somewhere? How can I factor $u^2-5u+4x$?

Answer 1

$$x^5-5x^3+4x=x(x^4-5x^2+4)$$$$=x(x^4-4x^2-x^2+4)$$$$=x(x^2(x^2-4)-(x^2-4))$$$$=x((x^2-4)(x^2-1))$$ $$=x(x+2)(x-2)(x+1)(x-1)$$

Answer 2

You have to factor $u^2 - 5u+4$, not $u^2 - 5u +4x$. Since $1\cdot 4 = 4$ and $1+ 4 = 5$, $u^2-5u+4=(u-4)(u-1)=(x^2-4)(x^2-1)$