How do you factor 4x^5-16x^3+12x ?

Nov 26, 2017

$4 {x}^{5} - 16 {x}^{3} + 12 x = 4 x \left(x - 1\right) \left(x + 1\right) \left(x - \sqrt{3}\right) \left(x + \sqrt{3}\right)$

Explanation:

The difference of squares identity can be written:

${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$

We will use this a couple of times, but first separate out the common factor $4 x$...

$4 {x}^{5} - 16 {x}^{3} + 12 x = 4 x \left({x}^{4} - 4 {x}^{2} + 3\right)$

$\textcolor{w h i t e}{4 {x}^{5} - 16 {x}^{3} + 12 x} = 4 x \left({\left({x}^{2}\right)}^{2} - 4 \left({x}^{2}\right) + 3\right)$

$\textcolor{w h i t e}{4 {x}^{5} - 16 {x}^{3} + 12 x} = 4 x \left({x}^{2} - 1\right) \left({x}^{2} - 3\right)$

$\textcolor{w h i t e}{4 {x}^{5} - 16 {x}^{3} + 12 x} = 4 x \left({x}^{2} - {1}^{2}\right) \left({x}^{2} - {\left(\sqrt{3}\right)}^{2}\right)$

$\textcolor{w h i t e}{4 {x}^{5} - 16 {x}^{3} + 12 x} = 4 x \left(x - 1\right) \left(x + 1\right) \left(x - \sqrt{3}\right) \left(x + \sqrt{3}\right)$