How do you factor #x^6 + 8#?

3 Answers
Apr 17, 2015

You express #8# as a power of #2#.

#8 = underbrace(2 * 2 * 2)_(color(blue)("3 times")) = 2^(color(blue)(3))#

Now your expression becomes

#E = x^(6) + 2^(3)#

You can express #x^(6)# as a cube by using

#x^(6) = x^(2 * 3) = (x^(2))^(3)#

Now you have a sum of two cubes, which you can factor according to the formula

#a^3 + b^3 = (a + b)(a^2 -ab + b^2)#

So, your expression will become

#E = (x^2)^3 + 2^3 = (x^2 + 2)((x^2)^2 - (x^2)2 + 2^2)#

#E = (x^2 + 2)(x^4 - 2x^2 + 4)#

Apr 17, 2015

This polynomial can be seen as a sum of cubes.

Remember the rule of factorize a sum of cubes:

#a^3+b^3=(a+b)(a^2-ab+b^2)#.

#x^6+8=(x^2)^3+2^3=(x^2+2)(x^4-2x^2+4)=(1)#,

generally the second factor can't be further factored, but this happens only if it is a square polynomial, but... this is a 4th degree polynomial, so:

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

#=(x^2+2)^2-(sqrt6x)^2=#

#=(x^2+2+sqrt6x)(x^2+2-sqrt6x)#,

and they can't be further factored.

Finally:

#(1)=(x^2+2)(x^2+2+sqrt6x)(x^2+2-sqrt6x)#.

Apr 17, 2015

#x^6# + 8 = (#x^2#+ 2)(#x^4#- 2#x^2# + 4)=(#x^2#+2)((#x^2#-1#)^2#+3)
Factors are: x + #sqrt(2)#i, x-#sqrt(2)#i, x-1+#sqrt(3)#i, x-1 - #sqrt(3)#i