How do you factor complete #x^3-8x^2+9x+18=0#?

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Answer 1 · 2015-02-05T03:36:23

We need to rewrite the polynomial equation so that we can properly factor it.

First rewrite #-8x^2# as #-9x^2+x^2#.

This gives us #x^3-9x^2+x^2+9x+18=0#

We now need to rewrite #9x# as #18x-9x#

This gives us #x^3-9x^2+18x+x^2-9x+18#

Now we are ready to factor.

#x(x^2-9x+18)+(x-6)(x-3)=0#
#x(x-6)(x-3)+(x-6)(x-3)=0#
#(x-6)(x-3)(x+1)=0#

Once the equation is factored we can set each factor equal to 0 to find the solutions, x-intercepts, to the polynomial equation.

#x-6=0#
#x=6#

#x-3=0#
#x=3#

#x+1=0#
#x=-1#