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Question 722968: (y^3-y^2-33y-21)/(y+5)
the ^ is suppose to mean its a exponent so y exponent 3 and y exponent 2
**Sorry about that..
Answer by 119078(26)   (Show Source):
You can put this solution on YOUR website! Start by getting the problem set up
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y+5|y^3-y^2-33y-21
Next, how many more y's would you need to multiply to y to get y^3? y^2. So that is the first part of your problem. You then multiply (y*y^2) and then 5*y^2 which gives you (y^3+5y^2). Subtract that to y^3-y^2
y^2
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y+5|y^3-y^2-33y-21
-(y^3+5y^2)
-6y^2-33y After getting your answer bring down -33y next to -6y^2.
What times y would give you -6y^2? -6y. (y*-6y)+(5*-6y)= (-6y^2-30y). Subtract (-6y^2-30y) to (-6y^2-33y).
y^2-6y
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y+5|y^3-y^2-33y-21
-(y^3+5y^2)
0 -6y^2-33y
-(-6y^2-30y)
0 -3y-21 After subtracting bring -21 down besides -3y.
Now for the last go around, what times y gives you -3y? -3 just as you were thinking! so for the last time multiply -3 to y and then -3 to 5. This will give you (-3y-15), subtract that to (-3y-21).
y^2-6y-3
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y+5|y^3-y^2-33y-21
-(y^3+5y^2)
0 -6y^2-33y
-(-6y^2-30y)
0 -3y-21
-(-3y-15)
0 -6 Now with the -6 that is your remainder. To finish the problem just but it at the end of the quadratic formula.
Here is your complete answer: y^2-6y-3 R-6.
I hope my step by step will help you get threw more problems like this. Yes it takes a long time to get these kind of problems done but it is all worth it in the end, because then you don't have to do it anymore! Have fun with the rest of your math!
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