# How do you factor x^4-8x^2+16?

Mar 13, 2018

${\left({x}^{2} - 4\right)}^{2}$

#### Explanation:

Make a dumby variable, in this case I will use $u$.
Let $u = {x}^{2}$
You have ${x}^{4} - 8 {x}^{2} + 16$.
This is equivalent to ${\left({x}^{2}\right)}^{2} - 8 {x}^{2} + 16$.
You can now plug in $u$ wherever ${x}^{2}$ is.
You get ${u}^{2} - 8 u + 16$.
Now, you can factor this like a normal polynomial.
$\left(u - 4\right) \left(u - 4\right)$
or
${\left(u - 4\right)}^{2}$.
Finally, you plug in $u$ back into the expression.
The factored form will be ${\left({x}^{2} - 4\right)}^{2}$.

Mar 13, 2018

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

#### Explanation:

$= \left({x}^{4} - 8 {x}^{2} + 16\right)$
$= \left({x}^{2} - 4\right) \left({x}^{2} - 4\right)$
$= \left(x + 2\right) \left(x - 2\right) \left(x + 2\right) \left(x - 2\right)$
$= {\left(x + 2\right)}^{2} {\left(x - 2\right)}^{2}$

Mar 13, 2018

Notice how this is almost a simple quadratic trinomial. In fact, it can be thought of as a quadratic of ${x}^{2}$.

Let $\textcolor{b l u e}{u = {x}^{2}}$

${x}^{4} - 8 {x}^{2} + 16$

$= {\left({x}^{2}\right)}^{2} - 8 \left({x}^{2}\right) + 16$

$= {\textcolor{b l u e}{u}}^{2} - 8 \textcolor{b l u e}{u} + 16$

We can easily factor this quadratic.

$= {\left(\textcolor{b l u e}{u} - 4\right)}^{2}$

$= {\left(\textcolor{b l u e}{{x}^{2}} - 4\right)}^{2}$