Is it possible to factor #y=x^3+5x^2-14x #? If so, what are the factors?

1 Answer
Dec 15, 2015

Answer:

Yes; #x=-7, 0, 2#

Explanation:

Original Equation:
#y=x^3+5x^2-14x#

Factor out an #x#:
#y=x(x^2+5x-14)#

Factor inside the parentheses (factors of #AC# that add up to #B#):
––Factors of #-14# that add up to #5#

You should get #7# and #-2#

Now plug it back into the equation:
#y=x(x-2)(x+7)#

Your factors are:
#x#
#(x-2)#
#(x+7)#

To find #x#, set each factor equal to #0# and solve.
Your final answer should be:
#x=0#
#x=2#
#x=-7#