# How do you solve 0=4#x^3#-20#x#+16?

##### 1 Answer

Apr 30, 2016

#### Explanation:

First note that all of the terms are divisible by

#0 = 4x^3-20x+16 = 4(x^3-5x+4)#

Next note that the sum of the coefficients of the simplified cubic is zero. that is:

#x^3-5x+4 = (x-1)(x^2+x-4)#

The remaining quadratic factor is in the form

#x = (-b+-sqrt(b^2-4ac))/(2a) = (-1+-sqrt(1^2-(4*1*-4)))/(2*1)#

#=(-1+-sqrt(17))/2 = -1/2+-sqrt(17)/2#