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

1 Answer
Apr 30, 2016

Answer:

#{ (x=1), (x=-1/2+sqrt(17)/2), (x=-1/2-sqrt(17)/2) :}#

Explanation:

First note that all of the terms are divisible by #4#, so we can separate that out to get:

#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:#1-5+4 = 0#. Hence #x=1# is a zero and #(x-1)# a factor:

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

The remaining quadratic factor is in the form #ax^2+bx+c# with #a=1#, #b=1# and #c=-4#. This has zeros given by the quadratic formula:

#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#