# How do you factor 4y=x^3-4x^2-11x+30 ?

Sep 2, 2016

$4 y = \left(x - 2\right) \left(x - 5\right) \left(x + 3\right)$

#### Explanation:

By the rational roots theorem, any rational zeros of ${x}^{3} - 4 {x}^{2} - 11 x + 30$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $30$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 3 , \pm 5 , \pm 6 , \pm 10 , \pm 15 , \pm 30$

If $x = 2$ then we find:

${x}^{3} - 4 {x}^{2} - 11 x + 30 = \left(8\right) - 4 \left(4\right) - 11 \left(2\right) + 30 = 8 - 16 - 22 + 30 = 0$

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

${x}^{3} - 4 {x}^{2} - 11 x + 30 = \left(x - 2\right) \left({x}^{2} - 2 x - 15\right)$

To factor the remaining quadratic, find a pair of numbers with difference $2$ and product $15$. The pair $5 , 3$ works, hence:

${x}^{2} - 2 x - 15 = \left(x - 5\right) \left(x + 3\right)$

Putting it all together, we have:

$4 y = \left(x - 2\right) \left(x - 5\right) \left(x + 3\right)$