# How do you factor 36x^4 + 15x^2 + 4?

May 23, 2015

First notice that $36 {x}^{4} + 15 {x}^{2} + 4$ has no linear factors with real coefficients, because $36 {x}^{4} \ge 0$ and $15 {x}^{2} \ge 0$ so

$36 {x}^{4} + 15 {x}^{2} + 4 \ge 4 > 0$ for all $x \in \mathbb{R}$.

$36 {x}^{4} + 15 {x}^{2} + 4 = \left(6 {x}^{2} + a x + b\right) \left(6 {x}^{2} + c x + d\right)$

$= 36 {x}^{4} + 6 \left(a + c\right) {x}^{3} + \left(6 d + 6 b + a c\right) {x}^{2} + \left(a d + b c\right) x + b d$

Comparing the coefficients of ${x}^{3}$ we find $c = - a$ so this simplifies to:

$= 36 {x}^{4} + \left(6 d + 6 b - {a}^{2}\right) {x}^{2} + a \left(d - b\right) x + b d$

Looking at the coefficient of $x$, either $a = 0$ or $d = b$.

If $a = 0$ then $6 \left(b + d\right) = 15$ and $b d = 4$. Substituting $d = \frac{4}{b}$ in $6 \left(b + d\right) = 15$ we get:

$6 \left(b + \frac{4}{b}\right) = 15$, and hence $6 {b}^{2} - 15 b + 24 = 0$. This has no real solutions, since it's discriminant is $225 - 576 = - 351$.

How about the other possibility: $d = b$?
Then $b = d = \pm 2$ in order that $b d = 4$.

Looking at the coefficient of ${x}^{2}$ we have
$15 = \left(6 d + 6 b - {a}^{2}\right) = 12 b - {a}^{2}$, hence $b = 2$ and $a = \pm 3$.

So $36 {x}^{4} + 15 {x}^{2} + 4 = \left(6 {x}^{2} + 3 x + 2\right) \left(6 {x}^{2} - 3 x + 2\right)$