# What is the solution set for 3x^5-48x=0?

Aug 8, 2015

$0 , \pm 2 , \pm 2 i$

#### Explanation:

Note that this is polynomial an equation of the 5th degree, so it should have 5 solutions.

$3 {x}^{5} - 48 x = 0$
$\implies 3 x \left({x}^{4} - 16\right) = 0$
$\implies x \left({\left({x}^{2}\right)}^{2} - {4}^{2}\right) = 0$ (Dividing both sides by 3)
$\implies x \left({x}^{2} + 4\right) \left({x}^{2} - 4\right) = 0$ (Since ${x}^{2} - {y}^{2} = \left(x + y\right) \left(x - y\right)$)
$\implies x \left({x}^{2} - \left(- 4\right)\right) \left({x}^{2} - 4\right) = 0$ (*)
$\implies x \left({x}^{2} - \left(- 4\right)\right) \left({x}^{2} - 4\right) = 0$
$\implies x \left({x}^{2} - {\left(2 i\right)}^{2}\right) \left({x}^{2} - {2}^{2}\right) = 0$ ( ${i}^{2} = - 1$)
$\implies x \left(x + 2 i\right) \left(x - 2 i\right) \left(x + 2\right) \left(x - 2\right) = 0$
$\implies x = 0 , \pm 2 , \pm 2 i$

If you are not looking for complex roots, at the step marked (*), note that ${x}^{2} + 4$ is always positive for all real values of $x$, and thus divide by ${x}^{2} + 4$. Then you can continue in the exact same way as given.