Question #9bee2

1 Answer
Mar 6, 2016

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=x^4#, yielding the trinomial

#u^2+5u+4#

This can simply be factored into

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

Since #u=x^4#, this is equivalent to

#(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 #x^4-(-4)# 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)=2i#.

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^4-(-1)#, 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

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