# Question #9bee2

Mar 6, 2016

See explanation.

#### Explanation:

We want to factor the trinomial

${x}^{8} + 5 {x}^{4} + 4$

This may look intimidating, since it's an octic equation. However, this becomes much more simple and familiar if we set $u = {x}^{4}$, yielding the trinomial

${u}^{2} + 5 u + 4$

This can simply be factored into

$\left(u + 4\right) \left(u + 1\right)$

Since $u = {x}^{4}$, this is equivalent to

$\left({x}^{4} + 4\right) \left({x}^{4} + 1\right)$

We can continue by factoring these using imaginary numbers.

Both of these terms can be factored as differences of squares, which take the form

${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

You may be asking yourself, how can these be differences of squares when the terms are being added?

The remedy to which is that the expression also equals

$\left({x}^{4} - \left(- 4\right)\right) \left({x}^{4} - \left(- 1\right)\right)$

If we treat ${x}^{4} - \left(- 4\right)$ as ${a}^{2} - {b}^{2}$, we see that ${a}^{2} = {x}^{4}$ and ${b}^{2} = - 4$, so $a = \sqrt{{x}^{4}} = {x}^{2}$ and $b = \sqrt{- 4} = 2 i$.

Thus, the expression with a factored first term is

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

We use a very similar method to factor ${x}^{4} - \left(- 1\right)$, namely ${a}^{2} = {x}^{4}$ and ${b}^{2} = - 1$, so $a = \sqrt{{x}^{4}} = {x}^{2}$ and $b = \sqrt{- 1} = i$, hence the expression equals

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