# How do you find all the zeros of #f(x)=4x^3-4x^2-7x-2#?

##### 1 Answer

#### Answer:

#### Explanation:

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

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

#+-1/4, +-1/2, +-1, +-2#

We can quickly see that

We find:

#f(2) = 32-16-14-2 = 0#

So

#4x^3-4x^2-7x-2 = (x-2)(4x^2+4x+1)#

To find the remaining zeros I could just try the other rational values we identified and factor by the next one we find, but there's a shortcut:

Do you see the pattern

Did you know that

As a result, we can spot that

#4x^2+4x+1 = (2x+1)^2#

This shows us that

graph{4x^3-4x^2-7x-2 [-4.877, 5.123, -2.4, 2.6]}