Question

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

Answer 1

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

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

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 `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:

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

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

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

So `x=2` is a zero and `(x-2)` a factor:

`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 `18` which differ by `3`

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

`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 `2`, `-6` and `3`

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

Answer 2

In short, `x^3+x^2-24x+36` would factor into `y=(x+6)(x-2)(x-3)`.

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
`(x+6)` is one of the three factors of the polynomial, leaving you with

`(x+6)(x^2-5x+6)`

From here you would factor the `(x^2-5x+6)` part into `(x-2)(x-3)`. Then put everything together to get

`(x+6)(x-2)(x-3)`

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 `(x+6)(x-2)(x-3)`, and you should get `x^3+x^2-24x+36` as an answer. Hope this helps!