# Question #9bee2

##### 1 Answer

See explanation.

#### Explanation:

We want to factor the trinomial

#x^8+5x^4+4#

This may look intimidating, since it's an octic equation. However, this becomes much more simple and familiar if we set

#u^2+5u+4#

This can simply be factored into

#(u+4)(u+1)#

Since

#(x^4+4)(x^4+1)#

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=(a+b)(a-b)#

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

#(x^4-(-4))(x^4-(-1))#

If we treat

Thus, the expression with a factored first term is

#(x^2+2i)(x^2-2i)(x^4-(-1))#

We use a very similar method to factor

#(x^2+2i)(x^2-2i)(x^2+i)(x^2-i)#