# How do you find the zeros of #f(x) = 2x^3 − 5x^2 − 28x + 15#?

##### 1 Answer

Zeros:

#### Explanation:

By the rational root theorem, since *rational* zeros must be expressible in the form

That means that the only possible *rational* zeros of

#+-1/2, +-1, +-3/2, +-5/2, +-3, +-5, +-15/2, +-15#

Trying the first one we find:

#f(1/2) = 2/8-5/4-28/2+15 = (1-5-56+60)/4 = 0#

So

#2x^3-5x^2-28x+15#

#=(2x-1)(x^2-2x-15)#

To factor the remaining quadratic, find a pair of factors of

#x^2-2x-15 = (x-5)(x+3)#

So the two remaining zeros are