# How do you factor #3x^3 - 9x^2 - 54x + 120#?

##### 1 Answer

Aug 18, 2016

#### Answer:

#### Explanation:

By the rational roots theorem, any *rational* zeros of

That means that the only possible *rational* zeros are:

#+-1, +-2, +-4, +-5, +-8, +-10, +-20, +-40#

Trying each in turn, with

#x^3-3x^2-18x+40=8-3(4)-18(2)+40 = 8-12-36+40 = 0#

So

#x^3-3x^2-18x+40 = (x-2)(x^2-x-20)#

To factor the remaining quadratic, note that

Hence:

#x^2-x-20 = (x-5)(x+4)#

Putting it all together:

#3x^3-9x^2-54x+120 = 3(x+2)(x-5)(x+4)#