# Is it possible to factor y= 6x^3-9x+3 ? If so, what are the factors?

Jan 9, 2016

Yes:

$y = 6 {x}^{3} - 9 x + 3$

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

$= 6 \left(x - 1\right) \left(x - \frac{1}{2} - \frac{\sqrt{3}}{2}\right) \left(x - \frac{1}{2} + \frac{\sqrt{3}}{2}\right)$

#### Explanation:

The difference of squares identity can be written:

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

We use this later.

First separate out the common scalar factor $3$ to find:

$y = 6 {x}^{3} - 9 x + 3 = 3 \left(2 {x}^{3} - 3 x + 1\right)$

Next note that the sum of the coefficients is zero. That is $2 - 3 + 1 = 0$. So $x = 1$ is a zero and $\left(x - 1\right)$ is a factor:

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

We can factor the remaining quadratic expression by completing the square and using the difference of squares identity...

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

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

$= 2 \left({x}^{2} - x + \frac{1}{4} - \frac{3}{4}\right)$

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

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

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

Putting it all together:

$y = 6 {x}^{3} - 9 x + 3$

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

$= 6 \left(x - 1\right) \left(x - \frac{1}{2} - \frac{\sqrt{3}}{2}\right) \left(x - \frac{1}{2} + \frac{\sqrt{3}}{2}\right)$