# How do you find all the zeros of #x^3-4x^2-44x+96#?

##### 1 Answer

Aug 14, 2016

This cubic has zeros:

#### Explanation:

#f(x) = x^3-4x^2-44x+96#

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

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

#+-1, +-2, +-3, +-4, +-6, +-8, +-12, +-16, +-24, +-32, +-48, +-96#

Trying each in turn, we find:

#f(2) = 8-16-88+96 = 0#

So

#x^3-4x^2-44x+96 = (x-2)(x^2-2x-48)#

To factor the remaining quadratic find a pair of factors of

#x^2-2x-48 = (x-8)(x+6)#

Hence zeros: