How do you solve #2x^3 – 5x^2 – 3x + 2 = 2#?

1 Answer
Jan 21, 2016

Answer:

#x=-1/2,0,3#

Explanation:

Begin by subtracting #2# from both sides of the equation.

#2x^3-5x^2-3x=0#

Factor an #x# from the terms on the left hand side.

#x(2x^2-5x-3)=0#

The quadratic is factorable into

#x(2x+1)(x-3)=0#

Now that you have these three terms being equal to #0#, you know that only one of them has to equal #0# for the entire expression to equal #0#. Thus, the three solutions can be found by setting each term equal to #0#:

#x=0#

#2x+1=0" "=>" "x=-1/2#

#x-3=0" "=>" "x=3#