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

2 Answers
Feb 15, 2018

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]}

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!