How do you factor 3x^3 - 9x^2 - 54x + 120?

Aug 18, 2016

$3 {x}^{3} - 9 {x}^{2} - 54 x + 120 = 3 \left(x + 2\right) \left(x - 5\right) \left(x + 4\right)$

Explanation:

$3 {x}^{3} - 9 {x}^{2} - 54 x + 120 = 3 \left({x}^{3} - 3 {x}^{2} - 18 x + 40\right)$

By the rational roots theorem, any rational zeros of ${x}^{3} - 3 {x}^{2} - 18 x + 40$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $40$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 4 , \pm 5 , \pm 8 , \pm 10 , \pm 20 , \pm 40$

Trying each in turn, with $x = 2$ we get:

${x}^{3} - 3 {x}^{2} - 18 x + 40 = 8 - 3 \left(4\right) - 18 \left(2\right) + 40 = 8 - 12 - 36 + 40 = 0$

So $x = 2$ is a zero and $\left(x - 2\right)$ a factor:

${x}^{3} - 3 {x}^{2} - 18 x + 40 = \left(x - 2\right) \left({x}^{2} - x - 20\right)$

To factor the remaining quadratic, note that $5 - 4 = 1$ and $5 \cdot 4 = 20$.

Hence:

${x}^{2} - x - 20 = \left(x - 5\right) \left(x + 4\right)$

Putting it all together:

$3 {x}^{3} - 9 {x}^{2} - 54 x + 120 = 3 \left(x + 2\right) \left(x - 5\right) \left(x + 4\right)$