# How do you find all the zeros of #f(x)=x^4+4x^3-6x^2-36x-27#?

##### 1 Answer

#### Answer:

#-1# (with multiplicity#1# )

#3# (with multiplicity#1# )

#-3# (with multiplicity#2# )

#### Explanation:

#f(x) = x^4+4x^3-6x^2-36x-27#

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

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

#+-1, +-3, +-9, +-27#

We find:

#f(-1) = 1-4-6+36-27 = 0#

So

#x^4+4x^3-6x^2-36x-27 = (x+1)(x^3+3x^2-9x-27)#

In the remaining cubic, the ratio of the first and second terms is the same as that between the third and fourth terms. So this cubic will factor by grouping:

#x^3+3x^2-9x-27#

#=(x^3+3x^2)-(9x+27)#

#=x^2(x+3)-9(x+3)#

#=(x^2-9)(x+3)#

#=(x^2-3^2)(x+3)#

#=(x-3)(x+3)(x+3)#

So the remaining zeros are