# Question #634fa

Aug 27, 2016

${x}^{4} - 2 {x}^{3} + 2 x - 1 = {\left(x - 1\right)}^{3} \left(x + 1\right)$

#### Explanation:

We will make use of the fact that if $P \left(x\right)$ is a polynomial, then $\left(x - a\right)$ is a factor of $P \left(x\right)$ if and only if $P \left(a\right) = 0$, along with polynomial long division.

If we look for a root of ${x}^{4} - 2 {x}^{3} + 2 x - 1$ (that is, a value such that the expression evaluates to $0$), we find that ${1}^{4} - 2 {\left(1\right)}^{3} + 2 \left(1\right) - 1 = 0$. Thus, using the above fact, we know that $\left(x - 1\right)$ is a factor of ${x}^{4} - 2 {x}^{3} + 2 x - 1$. Dividing, we find:

$\frac{{x}^{4} - 2 {x}^{3} + 2 x - 1}{x - 1} = {x}^{3} - {x}^{2} - x + 1$

$\implies {x}^{4} - 2 {x}^{3} + 2 x - 1 = \left(x - 1\right) \left({x}^{3} - {x}^{2} - x + 1\right)$

Next, we will look to factor ${x}^{3} - {x}^{2} - x + 1$. While there are more general ways of factoring cubic expressions, we can still test for roots first. Doing so, we find that $1$ is still a root of the remaining cubic equation, and so we once again divide by $\left(x - 1\right)$:

$\frac{{x}^{3} - {x}^{2} - x + 1}{x - 1} = {x}^{2} - 1$

$\implies \left({x}^{3} - {x}^{2} - x + 1\right) = \left(x - 1\right) \left({x}^{2} - 1\right)$

For the quadratic equation ${x}^{2} - 1$, there are multiple ways of continuing. The easiest is to note that it is an example of the special product $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$ with $a = x$ and $b = 1$. We could also test for roots as we did above and divide. Either way, we will find that

${x}^{2} - 1 = \left(x + 1\right) \left(x - 1\right)$

Now that we have factored everything down to polynomials of degree $1$, we can put it all together to get our factored form:

${x}^{4} - 2 {x}^{3} + 2 x - 1 = \left(x - 1\right) \left({x}^{3} - {x}^{2} - x + 1\right)$

$= \left(x - 1\right) \left(x - 1\right) \left({x}^{2} - 1\right)$

$= \left(x - 1\right) \left(x - 1\right) \left(x + 1\right) \left(x - 1\right)$

$= {\left(x - 1\right)}^{3} \left(x + 1\right)$