# How do you factor x^4 - 13x^2 - 30?

Jun 13, 2015

${x}^{4} - 13 {x}^{2} - 30 = \left({x}^{2} - 15\right) \left({x}^{2} + 2\right) = \left(x - \sqrt{15}\right) \left(x + \sqrt{15}\right) \left({x}^{2} + 2\right)$

#### Explanation:

${x}^{4} - 13 {x}^{2} - 30$ is quadratic in ${x}^{2}$.

To factor it, first find a pair of factors of $30$ whose difference is $13$. Such a pair is $15$ and $2$.

That gives us:

${x}^{4} - 13 {x}^{2} - 30 = \left({x}^{2} - 15\right) \left({x}^{2} + 2\right)$

The first of these quadratic factors can be considered a difference of squares:

${x}^{2} - 15 = {x}^{2} - {\left(\sqrt{15}\right)}^{2} = \left(x - \sqrt{15}\right) \left(x + \sqrt{15}\right)$

using the difference of squares identity:

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

The second factor ${x}^{2} + 2$ has no linear factors with real coefficients. If you were allowed complex coefficients then you could write:

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

Jun 13, 2015

$\left(x + \sqrt{15}\right) \left(x - \sqrt{15}\right) \left({x}^{2} + 2\right)$ in $\mathbb{R}$

$\left(x + \sqrt{15}\right) \left(x - \sqrt{15}\right) \left(x + \sqrt{2} i\right) \left(x - \sqrt{2} i\right)$ in $\mathbb{C}$

(If you have never used $i$ or $\mathbb{C}$, just ignore this last answer)

#### Explanation:

I'll call $P \left(x\right)$ the polynomial.

We notice that $P \left(x\right) = Q \left({x}^{2}\right)$ for some $Q$ quadratic polynomial. (In fact, P(x) has only even powers of $x$)

let $X = {x}^{2}$

So the equation becomes:

${X}^{2} - 13 X - 30$, which I know how to factor:

$a , b = \frac{13 \pm \sqrt{169 + 120}}{2} = \frac{13 \pm 17}{2} \implies a = 15 , b = - 2$

(if you had eyes wide open, you could notice that $- 13 = - \left(15 - 2\right)$ and $- 30 = \left(- 2\right) 15$, so you could avoid tedious calculations)

i.e. $Q \left(X\right) = \left(X - 15\right) \left(X + 2\right)$

So we factored the polynomial in

$P \left(x\right) = \left({x}^{2} - 15\right) \left({x}^{2} + 2\right)$

We notice that $\left({x}^{2} - 15\right) = \left(x + \sqrt{15}\right) \left(x - \sqrt{15}\right)$

If you work in $\mathbb{R}$, then $\left({x}^{2} + 2\right)$ is irreducible (it has no roots)

So $P \left(x\right) = \left(x + \sqrt{15}\right) \left(x - \sqrt{15}\right) \left({x}^{2} + 2\right)$

If you work in $\mathbb{C}$, ${x}^{2} + 2 = \left(x + \sqrt{2} i\right) \left(x - \sqrt{2} i\right)$

So $P \left(x\right) = \left(x + \sqrt{15}\right) \left(x - \sqrt{15}\right) \left(x + \sqrt{2} i\right) \left(x - \sqrt{2} i\right)$

(If you have never used $i$ or $\mathbb{C}$, just ignore this last two sentences)