# How do you factor the expression 3x - 9x^3?

Dec 28, 2015

$3 x - 9 {x}^{3}$

$= 3 x \left(1 - 3 {x}^{2}\right)$

$= 3 x \left(1 - \sqrt{3} x\right) \left(1 + \sqrt{3} x\right)$

$= x \left(\sqrt{3} - 3 x\right) \left(\sqrt{3} + 3 x\right)$

#### Explanation:

We can use the difference of squares identity:

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

to factor with irrational coefficients.

First notice that both terms are divisible by $3 x$, so separate that out as a factor:

$3 x - 9 {x}^{3}$

$= 3 x \left(1 - 3 {x}^{2}\right)$

$= 3 x \left({1}^{2} - {\left(\sqrt{3} x\right)}^{2}\right)$

$= 3 x \left(1 - \sqrt{3} x\right) \left(1 + \sqrt{3} x\right)$

Or if you prefer, split that factor of $3$ between the two binomials:

$3 x - 9 {x}^{3}$

$= x \left(3 - 9 {x}^{2}\right)$

$= x \left({\left(\sqrt{3}\right)}^{2} - {\left(3 x\right)}^{2}\right)$

$= x \left(\sqrt{3} - 3 x\right) \left(\sqrt{3} + 3 x\right)$