How do you factor x^6 + 8?

Apr 17, 2015

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

$8 = {\underbrace{2 \cdot 2 \cdot 2}}_{\textcolor{b l u e}{\text{3 times}}} = {2}^{\textcolor{b l u e}{3}}$

$E = {x}^{6} + {2}^{3}$

You can express ${x}^{6}$ as a cube by using

${x}^{6} = {x}^{2 \cdot 3} = {\left({x}^{2}\right)}^{3}$

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

${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

$E = {\left({x}^{2}\right)}^{3} + {2}^{3} = \left({x}^{2} + 2\right) \left({\left({x}^{2}\right)}^{2} - \left({x}^{2}\right) 2 + {2}^{2}\right)$

$E = \left({x}^{2} + 2\right) \left({x}^{4} - 2 {x}^{2} + 4\right)$

Apr 17, 2015

This polynomial can be seen as a sum of cubes.

Remember the rule of factorize a sum of cubes:

${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$.

${x}^{6} + 8 = {\left({x}^{2}\right)}^{3} + {2}^{3} = \left({x}^{2} + 2\right) \left({x}^{4} - 2 {x}^{2} + 4\right) = \left(1\right)$,

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:

${x}^{4} - 2 {x}^{2} + 4 = {x}^{4} + 4 {x}^{2} - 6 {x}^{2} + 4 = {x}^{4} + 4 {x}^{2} + 4 - 6 {x}^{2} =$

$= {\left({x}^{2} + 2\right)}^{2} - {\left(\sqrt{6} x\right)}^{2} =$

$= \left({x}^{2} + 2 + \sqrt{6} x\right) \left({x}^{2} + 2 - \sqrt{6} x\right)$,

and they can't be further factored.

Finally:

$\left(1\right) = \left({x}^{2} + 2\right) \left({x}^{2} + 2 + \sqrt{6} x\right) \left({x}^{2} + 2 - \sqrt{6} x\right)$.

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