# How do you find all the zeros of # 2x^3+9x^2+6x-8#?

##### 1 Answer

Aug 13, 2016

#### Answer:

This cubic has zeros

#### Explanation:

#f(x) = 2x^3+9x^2+6x-8#

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

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

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

We find:

#f(-2) = 2(-8)+9(4)+6(-2)-8 = -16+36-12-8 = 0#

So

#2x^3+9x^2+6x-8=(x+2)(2x^2+5x-4)#

The remaining quadratic is in the form

We can solve this using the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#=(-5+-sqrt(5^2-4(2)(-4)))/(2*2)#

#=1/4(-5+-sqrt(25+32))#

#=1/4(-5+-sqrt(57))#