# How do you factor 9x³+6x²-3x?

Jan 8, 2016

Use factor theorem, then long divide or compare coefficients to find the other factors.
Answer: $\left(9 {x}^{3} + 6 {x}^{2} - 3 x\right) = 3 x \left(x + 1\right) \left(3 x - 1\right)$

#### Explanation:

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

Using factor theorem:
($1$ and $- 1$ are typically good to try first)

$f \left(- 1\right) = - 9 + 6 + 3 = 0$

So $- 1$ is a root.
So $\left(x + 1\right)$ is a factor.

Here you have two options. Either long divide, or set up an equation and compare coefficients. The latter is more reliable in general, but algebraic long division is quicker as long as long as you don't get a remainder. If you get a remainder, abort, go the other way. I will show both methods. (ish).

Do algebraic long division : (so hard to explain in text, and layout isn't friendly for this either. If you don't know how to do algebraic long division, I would recommend looking for a video on it.)

$\frac{9 {x}^{3} + 6 {x}^{2} - 3 x}{x + 1} = 9 {x}^{2} - 3 x$ (no remainder)
So

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

OR

Do coefficient comparison:

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

${x}^{3}$ terms: $9 = a$
${x}^{2}$ terms: $6 = a + b$ $\implies$ $b = - 3$
$x$ terms: $- 3 = c + b$ $\implies$ $c = 0$

So:
$9 {x}^{3} + 6 {x}^{2} - 3 x = \left(x + 1\right) \left(9 {x}^{2} - 3 x\right)$ as before.
$\left(9 {x}^{3} + 6 {x}^{2} - 3 x\right) = 3 x \left(x + 1\right) \left(3 x - 1\right)$