# How do you find the roots of #x^3-x^2-34x-56=0#?

##### 1 Answer

#### Answer:

This cubic equation has roots

#### Explanation:

#f(x) = x^3-x^2-34x-56#

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

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

#+-1, +-2, +-4, +-7, +-8, +-14, +-28, +-56#

In addition, note that the pattern of signs of the coefficients of the terms is

If

#f(7) = 7^3-7^2-34(7)-56 = 343-49-238-56 = 0#

So

#x^3-x^2-34x-56 = (x-7)(x^2+6x+8)#

We can factor the remaining quadratic by finding a pair of numbers with sum

#x^2+6x+8 = (x+2)(x+4)#

Putting it all together:

#x^3-x^2-34x-56 = (x-7)(x+2)(x+4)#

with zeros