SOLUTION: Solve with Long Divison! THANK YOU:) (3c^5 + 5c^4 + c + 5) / (c+2)

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Solve with Long Divison! THANK YOU:) (3c^5 + 5c^4 + c + 5) / (c+2)      Log On


   

Question 162597: Solve with Long Divison! THANK YOU:)
(3c^5 + 5c^4 + c + 5) / (c+2)
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The 1st term is 3c%5E4
3c%5E4%2A%28c+%2B+2%29+=+3c%5E5+%2B+6c%5E4
Subtracting this from 3c%5E5+%2B+5c%5E4, I get
-c%5E4
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The 2nd term is -c%5E3
-c%5E3%2A%28c+%2B+2%29+=+-c%5E4+-+2c%5E3
Subtracting this from -c%5E4+%2B+c, I get
2c%5E3+%2B+c
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The 3rd term is 2c%5E2
2c%5E2%2A%28c+%2B+2%29+=+2c%5E3+%2B+4c%5E2
Subtracting this from 2c%5E3+%2B+c, I get
-4c%5E2+%2B+c
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The 4th term is -4c
-4c%2A%28c+%2B+2%29+=+-4c%5E2+-+8c
Subtracting this from -4c%5E2+%2B+c, I get
9c
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The 5th term is 9
9%2A%28c+%2B+2%29+=+9c+%2B+18
Subtracting this from 9c+%2B+5, I get
-13
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The answer is 3c%5E4+-+c%5E3+%2B+2c%5E2+-+4c+%2B+9+-+13%2F%28c%2B2%29
The last term is the remainder
Multiply this bt c%2B2 and see if you get
3c%5E5+%2B+5c%5E4+%2B+c+%2B+5
I did and it checked OK