# What are all the possible rational zeros for #f(x)=8x^3-2x+24# and how do you find all zeros?

##### 1 Answer

#x_1 = -3/2#

#x_(2,3) = 3/4+-sqrt(23)/4i#

#### Explanation:

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

So the only possible *rational* zeros of

#+-1/4, +-1/2, +-3/4, +-1, +-3/2, +-2, +-3, +-4, +-6, +-12#

By Descartes' rule of signs we can tell that

So let's try the negative possibilities first:

#f(-1/4) = 2(4(-1/64)-(-1/4)+12) = 2(-1/16+1/4+12) != 0#

#f(-1/2) = 2(4(-1/8)-(-1/2)+12) = 2(-1/2+1/2+12) = 24 != 0#

#f(-3/4) = 2(4(-27/64)-(-3/4)+12) = 2(-27/16+3/4+12) != 0#

#f(-1) = 2(-4+1+12) = 18 != 0#

#f(-3/2) = 2(4(-27/8)-(-3/2)+12) = 2(-27/2+3/2+24/2) = 0#

So

#4x^3-x+12 = (2x+3)(2x^2-3x+4)#

The remaining quadratic factor has zeros given by the quadratic formula:

#x = (3+-sqrt((-3)^2-4(2)(4)))/(2*2)#

#=(3+-sqrt(9-32))/4#

#=(3+-sqrt(-23))/4#

#=(3+-sqrt(23)i)/4#

#=3/4+-sqrt(23)/4i#