In this case I would use the method of synthetic substitution, but since I never learned this technique 'til this year in Precalculus, let's stick with long division:
1) Write out the problem like an elementary division problem, with the bar over the 4m^2-9m^2...+7 and the m-3 to the side.
2) For the beginning, calculate how many times m (not m-3, just m) goes into 4m^3. In simpler terms, what would you multiply m by to get 4m^3? You would get 4m^2, so write that on the bar above.
3) Like regular division, write the product of m-3 and 4m^2 underneath the problem, drop another set of the equation down (just ONE expression, which, in this case would be the -9m^2 next). So far, you should have 4m^2 + 12m^2 written underneath to subtract from the first two terms in the dividend.
4) Repeat the process until you can go no further, which is when you reach a number without any variables. This (divided by m-3) will be your remainder.
I hope this makes sense....it'd be easier to show if I could show in person. Hope this helps! :)
How would you solve this? I don't even know how to begin this... I know the answer is 4m^2 + 3m + 19 + (64/m-3) but I want to know how to do it incase its on the test.
4m^3 - 9m^2 + 10m + 7 / m-3
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1 answer