Question 134593
Use polynomial long division to divide {{{x+3}}} into {{{x^3+2x^2-5x-6}}}


{{{x}}} goes into {{{x^3}}} {{{x^2}}} times, so {{{x^2}}} is the first term of the quotient.


{{{x^2}}} times {{{x+3}}} is {{{x^3+3x^2}}}.


Subtract {{{x^3+3x^2}}} from {{{x^3+2x^2}}} and get {{{-x^2}}}


Bring down the {{{-5x}}} to form {{{-x^2-5x}}}


{{{x}}} goes into {{{-x^2}}} {{{-x}}} times, so {{{-x}}} becomes the second term of the quotient.


{{{-x}}} times {{{x+3}}} is {{{-x^2-3x}}}


Subtract {{{-x^2-3x}}} from {{{-x^2-5x}}} and get {{{-2x}}}


Bring down the {{{-6}}} to form {{{-2x-6}}}


{{{x}}} goes into {{{-2x}}} {{{-2}}} times, so {{{-2}}} becomes the third term of the quotient.


{{{-2}}} times {{{x+3}}} is {{{-2x-6}}}


Subtract {{{-2x-6}}} from {{{-2x-6}}} and get 0.  So you have verified that {{{x+3}}} is a factor.  Now put the quotient together:


{{{x^2-x-2}}}


{{{-2*1=-2}}} and {{{-2+1=-1}}}, so the trinomial quotient factors to:


{{{(x-2)(x-1)}}}


Therefore:  {{{x^3+2x^2-5x-6=(x+3)(x-2)(x-1)}}}