# How do you factor x^4 - x^2 - 20?

Jun 25, 2016

$\left(x - 2\right) \left(x + 2\right) \left(x - i \sqrt{5}\right) \left(x + i \sqrt{5}\right)$

#### Explanation:

create new variable equal to ${x}^{2}$. Call it y.

Your equation now is ${y}^{2} - y - 20$ whose roots are $y = 4$ and $y = - 5$
$y = 4$ is ok. That is $x = \pm 2$. $y = - 5$ leads to imaginary roots $x = \pm i \sqrt{5}$

Jun 25, 2016

${x}^{4} - {x}^{2} - 20 = \left({x}^{2} + 4\right) \left({x}^{2} - 5\right)$
= $\left(x + \sqrt{5}\right) \left(x - \sqrt{5}\right) \left({x}^{2} + 4\right)$
= $\left(x + \sqrt{5}\right) \left(x - \sqrt{5}\right) \left(x + 2 i\right) \left(x - 2 i\right)$

#### Explanation:

${x}^{4} - {x}^{2} - 20$

= ${x}^{4} - 5 {x}^{2} + 4 {x}^{2} - 20$

= ${x}^{2} \left({x}^{2} - 5\right) + 4 \left({x}^{2} - 5\right)$

= $\left({x}^{2} + 4\right) \left({x}^{2} - 5\right)$

These are factors with rational coefficients. If we are allowed to have monomials with irrational coefficients too, ${x}^{2} - 5 = \left(x + \sqrt{5}\right) \left(x - \sqrt{5}\right)$ and

${x}^{4} - {x}^{2} - 20 = \left(x + \sqrt{5}\right) \left(x - \sqrt{5}\right) \left({x}^{2} + 4\right)$

If we also allow monomials with complex numbers, then factors of ${x}^{2} + 4$ are $\left(x + 2 i\right) \left(x - 2 i\right)$ and

${x}^{4} - {x}^{2} - 20 = \left(x + \sqrt{5}\right) \left(x - \sqrt{5}\right) \left(x + 2 i\right) \left(x - 2 i\right)$

Jun 25, 2016

$E x p . = \left({x}^{2} - 5\right) \left({x}^{2} + 4\right) .$

#### Explanation:

Expression $= {x}^{4} - {x}^{2} - 20.$

Note that $5 \times 4 = 20 , 5 - 4 = 1$

$\therefore E x p . = {x}^{4} - 5 {x}^{2} + 4 {x}^{2} - 20 = {x}^{2} \left({x}^{2} - 5\right) + 4 \left({x}^{2} - 5\right) = \left({x}^{2} - 5\right) \left({x}^{2} + 4\right) .$

If, at the first glance, it may not strike that the given Exp. is a qudr. poly, of ${x}^{2}$, then, we may use substitution ${x}^{2} = t$ in it and see that it is ${t}^{2} - t - 20 = \left(t - 5\right) \left(t + 4\right) = \left({x}^{2} - 5\right) \left({x}^{2} + 4\right) ,$ as before!.