# How do you factor 3u^4 - 19u^2 + 28?

Jun 4, 2015

$3 {u}^{4} - 19 {u}^{2} + 28$

$= 3 {\left({u}^{2}\right)}^{2} - 19 \left({u}^{2}\right) + 28$

$= \left(3 {u}^{2} - 7\right) \left({u}^{2} - 4\right)$

$= \left(3 {u}^{2} - 7\right) \left(u - 2\right) \left(u + 2\right)$

If we allow irrational factors:

$= \left(\sqrt{3} u - \sqrt{7}\right) \left(\sqrt{3} u + \sqrt{7}\right) \left(u - 2\right) \left(u + 2\right)$

To find the factors for the factorization into $\left(3 {u}^{2} - 7\right) \left({u}^{2} - 4\right)$,
I basically tried a few combinations of the form $\left(3 {u}^{2} - a\right) \left({u}^{2} - b\right)$ with $a b = 28$.

Having recognised that the quartic is a quadratic in ${u}^{2}$ I could have used any of the normal methods (quadratic formula, AC Method, etc.) for factoring quadratics, but it was simple enough to try a couple of guesses.

The factorizations into linear factors are a couple of examples of the general difference of squares identity:

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