How do you solve #x^3+x^2-16x-16=0#?

1 Answer
Sep 26, 2017

Answer:

#x in { -1, -4, +4}#

Explanation:

Given
#color(white)("XXX")x^3+x^2-16x-16=0#

By observation we can see that #x=-1# is an obvious solution.
That is #(x+1)# is a factor of the left side of the given equation.

Using polynomial long division or synthetic division we get:
#color(white)("XXX")(x+1)(x^2-16)=0#

#(x^2-16)# is obviously the difference of squares with
#color(white)("XXX")(x^2-16)=(x+4)(x-4)#

So we have
#color(white)("XXX")(x+1)(x+4)(x-4)=0#
which implies
#(x+1)=0color(white)("xxx")rarrcolor(white)("xxx")x=-1#
or
#(x+4)=0color(white)("xxx")rarrcolor(white)("xxx")x=-4#
or
#(x-4)=0color(white)("xxx")rarrcolor(white)("xxx")x=4#