Question 124827
What you need to do is polynomial long division.  The process is difficult to render on this system, so follow along carefully.


The first thing that you have to recognize is that there is no {{{m}}} term in your divisor.  So you have to re-write the divisor to be {{{m^2+0m-7}}}


Divide the lead term of the divisor into the lead term of the dividend.  {{{m^2}}} goes into {{{m^3}}} {{{m}}} times.  Write down {{{m}}} as the first term of your quotient.


Multiply the first term of your quotient, {{{m}}} by each term of the divisior and put your results under the dividend, lining up like terms.  {{{m}}} times {{{-7}}} is {{{-7m}}}, {{{m}}} times {{{0m}}} is {{{0m^2}}}, and finally {{{m}}} times {{{m^2}}} is {{{m^3}}}.  This is your first partial product.


Subtract the first partial product from the dividend, just like when you are doing regular long division, except subtract the two polynomials term-by-term.


{{{-7m-(-7m)=0m}}}
{{{3m^2-0m^2=3m^2}}}
{{{m^3-m^3=0}}}

That leaves you with {{{3m^2+0m-21}}} as a partial dividend.  Divide the lead term of your divisor into the lead term of this partial dividend.  {{{m^2}}} goes into {{{3m^2}}} {{{3}}} times.  Put a plus {{{+3}}} in your quotient.


Now multiply this {{{3}}} times the divisor, term by term.  {{{3*(-7)=-21}}}, {{{3*0m=0m}}}, and {{{3*m^2=3m^2}}}.  Put this entire expression beneath your partial dividend, lining up like terms.


Now subtract.  If you have done everything correctly, you should have a result of 0.  That means that {{{m^2-7}}} divides, specifically {{{m+3}}} times, into {{{m^3+3m^2-7m-21}}}, with no remainder.


Now we can say that {{{(m+3)(m^2-7)=m^3+3m^2-7m-21}}}.  Multiply it out to check the answer.