How do you find all zeroes for #f(x)=2x^3-3x^2+1#?

1 Answer
Feb 1, 2017

Answer:

The zeros of #f(x)# are: #1#, #1#, #-1/2#

Explanation:

Given:

#f(x) = 2x^3-3x^2+1#

First note that the sum of the coefficients is #0#. That is:

#2-3+1 = 0#

Hence #f(1) = 0# and #(x-1)# is a factor:

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

Note that the sum of the coefficients of the remaining quadratic is also zero:

#2-1-1 = 0#

So #x=1# is a zero again and #(x-1)# a factor again:

#2x^2-x-1 = (x-1)(2x+1)#

From the last linear factor we can see that the remaining zero is #x=-1/2#