Question 133347
Do you want to find the zeros? If so, I'll do the first two for you



{{{x^3-5x^2-14x=0}}} Start with the given equation



{{{x(x^2-5x-14)=0}}} Factor out the common term "x"



{{{x(x-7)(x+2)=0}}} Factor the inner expression (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)




Now set each factor equal to zero:

{{{x=0}}}, {{{x-7=0}}} or {{{x+2=0}}}


{{{x=0}}}, {{{x=7}}} or {{{x=-2}}}   Now solve for x in each case



So our answer is 

 {{{x=0}}}, {{{x=7}}} or {{{x=-2}}}



Notice if we graph {{{y=x^3-5x^2-14x}}}  we can see that the roots are {{{x=0}}}, {{{x=7}}} and {{{x=-2}}}. So this visually verifies our answer.



{{{ graph(500,500,-10,10,-10,10,0, x^3-5x^2-14x) }}}



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# 2



{{{x^3+4x^2+9x+36=0}}} Start with the given equation


{{{(x^3+4x^2)+(9x+36)=0}}} Group like terms



{{{x^2(x+4)+9(x+4)=0}}} Factor out the GCF {{{x^2}}} out of the first group. Factor out the GCF {{{9}}} out of the second group



{{{(x^2+9)(x+4)=0}}} Since we have the common term {{{x+4}}}, we can combine like terms





Now set each factor equal to zero:


{{{x^2+9=0}}} or  {{{x+4=0}}} 


Now solve for x for each factor:


{{{x=0+-sqrt(-9)}}} or  {{{x=-4}}} 



{{{x=3i}}}, {{{x=-3i}}} or  {{{x=-4}}} 



So our solutions are 


{{{x=-4}}}, {{{x=3i}}} or {{{x=-3i}}}