# What are all the possible rational zeros for #f(x)=6x^3-x^2-13x+8# and how do you find all zeros?

##### 1 Answer

This cubic has "possible" *rational* zeros:

#+-1/6, +-1/3, +-2/3, +-1, +-4/3, +-2, +-8/3, +-4, +-8#

and actual zeros:

#1" "# and#" "-5/12+-sqrt(217)/12#

#### Explanation:

#f(x) = 6x^3-x^2-13x+8#

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

That means that the only possible rational zeros are:

#+-1/6, +-1/3, +-2/3, +-1, +-4/3, +-2, +-8/3, +-4, +-8#

There is a shortcut in that the sum of the coefficients is

That is:

#6-1-13+8 = 0#

So

#6x^3-x^2-13x+8 = (x-1)(6x^2+5x-8)#

The remaining quadratic factor is in the form

This has discriminant

#Delta = b^2-4ac = 5^2-4(6)(-8) = 25+192 = 217#

Since

We can find the zeros using the quadratic formula:

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

#color(white)(x) = (-b+-sqrt(Delta))/(2a)#

#color(white)(x) = (-5+-sqrt(217))/12#

#color(white)(x) = -5/12+-sqrt(217)/12#