How do you factor #x^ { 8} + 2x ^ { 6} - x ^ { 4} - 2x ^ { 2}#?

1 Answer
George C.
Jan 26, 2017

#x^8+2x^6-x^4-2x^2 = x^2(x-1)(x+1)(x^2+1)(x^2+2)#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

We use this a couple of times below.

Given:

#x^8+2x^6-x^4-2x^2#

Note that the ratio of the first and second terms is the same as that between the third and fourth terms. So this octic quadrinomial will factor by grouping. In addition all of the terms are divisible by #x^2#, so we can separate that out as a factor first...

#x^8+2x^6-x^4-2x^2 = x^2(x^6+2x^4-x^2-2)#

#color(white)(x^8+2x^6-x^4-2x^2) = x^2((x^6+2x^4)-(x^2+2))#

#color(white)(x^8+2x^6-x^4-2x^2) = x^2(x^4(x^2+2)-1(x^2+2))#

#color(white)(x^8+2x^6-x^4-2x^2) = x^2(x^4-1)(x^2+2)#

#color(white)(x^8+2x^6-x^4-2x^2) = x^2((x^2)^2-1^2)(x^2+2)#

#color(white)(x^8+2x^6-x^4-2x^2) = x^2(x^2-1)(x^2+1)(x^2+2)#

#color(white)(x^8+2x^6-x^4-2x^2) = x^2(x^2-1^2)(x^2+1)(x^2+2)#

#color(white)(x^8+2x^6-x^4-2x^2) = x^2(x-1)(x+1)(x^2+1)(x^2+2)#

The remaining quadratic factors are strictly positive for any Real value of #x#, and hence cannot be broken down into linear factors with Real coefficients.

Related questions
Impact of this question
950 views around the world
You can reuse this answer
Creative Commons License