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

##### 2 Answers

with zeros

#### Explanation:

Given:

#f(x) = x^3+x^2-24x+36#

By the rational roots theorem, any rational root of this cubic is expressible in the form

So the only possible rational zeros are:

#+-1, +-2, +-3, +-4, +-6, +-9, +-12, +-18, +-36#

Looking at the sizes of the coefficients and their signs, I think I'll try

#f(2) = 8+4-48+36 = 0#

So

#x^3+x^2-24x+36 = (x-2)(x^2+3x-18)#

To factor the remaining quadratic, we can find a pair of factors of

The pair

#x^2+3x-18 = (x+6)(x-3)#

Putting it all together:

#x^3+x^2-24x+36 = (x-2)(x+6)(x-3)#

with zeros

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

In short,

#### Explanation:

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

First, start by finding all of the factors of 36:

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

Then, by trial and error, you would eventually find that

From here you would factor the

Thus, the zeros of the polynomial are

P.S. To verify this, you can just expand the factored form, i.e.

the