# How do you find all the zeros of #x^4 - 4x^3 - 20x^2 + 48x#?

##### 1 Answer

#### Explanation:

Firstly, since all of the factors are divisible by

Dividing through by

#f(x) = x^3-4x^2-20x+48#

By the rational root theorem, any rational zeros of

That means that the only possible rational zeros are:

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

We find

So

#x^3-4x^2-20x+48 = (x-2)(x^2-2x-24)#

To factor the remaining quadratic, note that

#x^2-2x-24 = (x-6)(x+4)#

Putting it all together:

#x^4-4x^3-20x^2+48x = x(x-2)(x-6)(x+4)#