Question 470600
First, we must factor.





{{{x^3+2x^2-16x-32}}} Start with the given expression



{{{(x^3+2x^2)+(-16x-32)}}} Group like terms



{{{x^2(x+2)-16(x+2)}}} Factor out the GCF {{{x^2}}} out of the first group. Factor out the GCF {{{-16}}} out of the second group



{{{(x^2-16)(x+2)}}} Since we have the common term {{{x+2}}}, we can combine like terms




{{{(x+4)(x-4)(x+2)}}} Now factor {{{x^2-16}}} to get {{{(x+4)(x-4)}}} (this a difference of squares)



So {{{x^3+2x^2-16x-32}}} factors to {{{(x+4)(x-4)(x+2)}}}



In other words, {{{x^3+2x^2-16x-32=(x+4)(x-4)(x+2)}}}


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So this essentially means that {{{x^3+2x^2-16x-32=0}}} becomes {{{(x+4)(x-4)(x+2)=0}}}



Now we move onto solving {{{(x+4)(x-4)(x+2)=0}}}




{{{(x+4)(x-4)(x+2)=0}}} Start with the given equation.



{{{x+4=0}}} or {{{x-4=0}}}, or {{{x+2=0}}} Use the zero product property



{{{x=-4}}} or {{{x=4}}}, or {{{x=-2}}} Solve for 'x' in each equation.




So the three solutions are {{{x=-4}}} or {{{x=4}}}, or {{{x=-2}}}