How do you factor #x^6+125#?
1 Answer
Explanation:
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
#(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