# How do you factor completely: 6x^4-9x^3-36x^2+54x?

Jul 12, 2015

Factor by grouping to get:

$6 {x}^{4} - 9 {x}^{3} - 36 {x}^{2} + 54 x$

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

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

#### Explanation:

First, note that every term is divisible by $3 x$, so

$6 {x}^{4} - 9 {x}^{3} - 36 {x}^{2} + 54 x = 3 x \left(2 {x}^{3} - 3 {x}^{2} - 12 x + 18\right)$

Next, to factor $2 {x}^{3} - 3 {x}^{2} - 12 x + 18$, notice that the ratio of the 1st and 2nd terms is the same as the ratio of the 3rd and 4th terms. So factoring by grouping will work:

$2 {x}^{3} - 3 {x}^{2} - 12 x + 18$

$= \left(2 {x}^{3} - 3 {x}^{2}\right) - \left(12 x - 18\right)$

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

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

Next, use the difference of squares identity:

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

with $a = x$ and $b = \sqrt{6}$ to get:

$\left({x}^{2} - 6\right) \left(2 x - 3\right)$

$= \left({x}^{2} - {\sqrt{6}}^{2}\right) \left(2 x - 3\right)$

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

Jul 12, 2015

The answer is $3 x \left(2 x - 3\right) \left(x + \sqrt{6}\right) \left(x - \sqrt{6}\right)$.

#### Explanation:

Factor by grouping.

$\left(6 {x}^{4} - 9 {x}^{3}\right) - \left(36 {x}^{2} - 54 x\right)$

Factor out the GCF for each group.

$3 {x}^{3} \left(2 x - 3\right) - 18 x \left(2 x - 3\right)$

Factor out $\left(2 x - 3\right)$.

$\left(3 {x}^{3} - 18 x\right) \left(2 x - 3\right)$

$\left(3 {x}^{3} - 18 x\right)$ can be factored further.

$3 x \left({x}^{2} - 6\right)$

Substitute it back into the previous factorization.

$3 x \left({x}^{2} - 6\right) \left(2 x - 3\right)$

$\left({x}^{2} - 6\right)$ can be factored further.

$\left(x + \sqrt{6}\right) \left(x - \sqrt{6}\right)$

Put all of the factors together.

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