SOLUTION: How do you solve for m^3+3m^2-7m-21 divided by m^2-7?

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Question 124827: How do you solve for m^3+3m^2-7m-21 divided by m^2-7?
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
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%5E2%2B0m-7

Divide the lead term of the divisor into the lead term of the dividend. m%5E2 goes into m%5E3 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%5E2, and finally m times m%5E2 is m%5E3. 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-%28-7m%29=0m
3m%5E2-0m%5E2=3m%5E2
m%5E3-m%5E3=0
That leaves you with 3m%5E2%2B0m-21 as a partial dividend. Divide the lead term of your divisor into the lead term of this partial dividend. m%5E2 goes into 3m%5E2 3 times. Put a plus %2B3 in your quotient.

Now multiply this 3 times the divisor, term by term. 3%2A%28-7%29=-21, 3%2A0m=0m, and 3%2Am%5E2=3m%5E2. 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%5E2-7 divides, specifically m%2B3 times, into m%5E3%2B3m%5E2-7m-21, with no remainder.

Now we can say that %28m%2B3%29%28m%5E2-7%29=m%5E3%2B3m%5E2-7m-21. Multiply it out to check the answer.