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

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]}

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!