# How do you factor x^3+x^2-24x+36 and what are its zeros?

Feb 15, 2018

${x}^{3} + {x}^{2} - 24 x + 36 = \left(x - 2\right) \left(x + 6\right) \left(x - 3\right)$

with zeros $2$, $- 6$ and $3$

#### Explanation:

Given:

$f \left(x\right) = {x}^{3} + {x}^{2} - 24 x + 36$

By the rational roots theorem, any rational root of this cubic is expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $36$ and $q$ a divisor of the coefficient $1$ of the leading term.

So the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 3 , \pm 4 , \pm 6 , \pm 9 , \pm 12 , \pm 18 , \pm 36$

Looking at the sizes of the coefficients and their signs, I think I'll try $x = 2$ first...

$f \left(2\right) = 8 + 4 - 48 + 36 = 0$

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

${x}^{3} + {x}^{2} - 24 x + 36 = \left(x - 2\right) \left({x}^{2} + 3 x - 18\right)$

To factor the remaining quadratic, we can find a pair of factors of $18$ which differ by $3$

The pair $6 , 3$ works, so we find:

${x}^{2} + 3 x - 18 = \left(x + 6\right) \left(x - 3\right)$

Putting it all together:

${x}^{3} + {x}^{2} - 24 x + 36 = \left(x - 2\right) \left(x + 6\right) \left(x - 3\right)$

with zeros $2$, $- 6$ and $3$

graph{x^3+x^2-24x+36 [-10, 10, -15, 105]}

Feb 15, 2018

In short, ${x}^{3} + {x}^{2} - 24 x + 36$ would factor into $y = \left(x + 6\right) \left(x - 2\right) \left(x - 3\right)$.

#### Explanation:

However, in order to get this you would need to use synthetic division.

First, start by finding all of the factors of 36: ±1, ±2, ±3, ±6, ±9, ±12, ±18, and ±36.

Then, construct a sideways "L" (as is always the case in synthetic division).

Then, by trial and error, you would eventually find that
$\left(x + 6\right)$ is one of the three factors of the polynomial, leaving you with

$\left(x + 6\right) \left({x}^{2} - 5 x + 6\right)$

From here you would factor the $\left({x}^{2} - 5 x + 6\right)$ part into $\left(x - 2\right) \left(x - 3\right)$. Then put everything together to get

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

Thus, the zeros of the polynomial are

$x = - 6 , x = 2 , x = 3$

P.S. To verify this, you can just expand the factored form, i.e.
the $\left(x + 6\right) \left(x - 2\right) \left(x - 3\right)$, and you should get ${x}^{3} + {x}^{2} - 24 x + 36$ as an answer. Hope this helps!