How do you factor #x^6+125#?

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Answer 1 · 2017-02-04T14:44:26

#x^6+125 = (x^2+5)(x^2-sqrt(15)x+5)(x^2+sqrt(15)x+5)#

The sum of cubes identity can be written:

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

Hence we find:

#x^6+125 = (x^2)^3+5^3#

#color(white)(x^6+125) = (x^2+5)((x^2)^2-5(x^2)+5^2)#

#color(white)(x^6+125) = (x^2+5)(x^4-5x^2+25)#

To factor the remaining quartic, note that:

#(a^2-kab+b^2)(a^2+kab+b^2) = a^4+(2-k^2)a^2b^2+b^4#

So with #a=x# and #b=sqrt(5)#, we find:

#(x^2-ksqrt(5)x+5)(x^2+ksqrt(5)x+5) = x^4+(2-k^2)5x^2+25#

Equating coefficients, we want:

#(2-k^2)5 = -5#

Hence:

#k = +-sqrt(3)#

So:

#x^4-5x^2+25 = (x^2-sqrt(3)sqrt(5)x+5)(x^2+sqrt(3)sqrt(5)x+5)#

#color(white)(x^4-5x^2+25) = (x^2-sqrt(15)x+5)(x^2+sqrt(15)x+5)#

Putting it all together:

#x^6+125 = (x^2+5)(x^2-sqrt(15)x+5)(x^2+sqrt(15)x+5)#

We can quickly determine that none of these quadratics has linear factors with real coefficients since #x^6+125# has no real zeros.