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