How to factor $56x^4+18x^2-8$?

Question

I've been trying to figure out how you solve this question but I just can't seem to understand how to factor

$$56x^4+18x^2-8$$

Answer 1

As others have hinted at, start by letting $t=x^2$. Then you get the following: $$ 56x^4+18x^2-8 = 56t^2+18t-8. $$ Now your job is to factor this: \begin{align} 56t^2+18t-8 &= 2(4t-1)(7t+4)\\[0.5em] &= 2(4x^2-1)(7x^2+4)\\[0.5em] &= 2(2x-1)(2x+1)(7x^2+4). \end{align} Thus, we see that $$ 56x^4+18x^2-8 = 2(2x-1)(2x+1)(7x^2+4). $$

Answer 2

Hint: Set $u=x^2$. Factor the resulting quadratic in $u$.

Answer 3

Hint: Let $y = x^2$ and then factor.

Answer 4

Another way is: Set $t=x^2$. So you have

$56t^2+18t-8=2(28t^2+9t-4)$

$\Delta=9^2+4\cdot28\cdot4=81+448=529$

$t_{1,2}=\frac{-9\pm \sqrt{529}}{56}=\frac{-9\pm 23}{56}$

$t_1=-\frac{32}{56}=-\frac{4}{7}$ $t_2=\frac{14}{56}=\frac{1}{4}$

so

$56(t-\frac{1}{4})(t+\frac{4}{7})$

$56(x^2-\frac{1}{4})(x^2+\frac{4}{7})=$

$56(x-\frac{1}{2})(x+\frac{1}{2})(x^2+\frac{4}{7})=$

$2(2x+1)(2x-1)(7x^2+4)$