# What are all the possible rational zeros for #f(x)=x^4-3x^3-6x^2+6x+8# and how do you find all zeros?

##### 1 Answer

#### Answer:

The "possible" rational zeros are:

The actual zeros are:

#### Explanation:

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

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

#+-1, +-2, +-4, +-8#

We find:

#f(-1) = 1+3-6-6+8 = 0#

So

#x^4-3x^3-6x^2+6x+8 = (x+1)(x^3-4x^2-2x+8)#

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

#x^3-4x^2-2x+8 = (x^3-4x^2)-(2x-8)#

#color(white)(x^3-4x^2-2x+8) = x^2(x-4)-2(x-4)#

#color(white)(x^3-4x^2-2x+8) = (x^2-2)(x-4)#

Finally we can factor the quadratic as a difference of squares:

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

So the zeros are: