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November 2nd, 2007, 06:52 PM   #1
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An Algebraic Long Division, yet the solution is very tricky!

Question: If a polynomial f(x) is divided by (x-1)^2, then it's remainder is 3x+2, and if the polynomial f(x) is divided by x+1, then the remainder is 3. Then, if the polynomial f(x) is divided by ((x-1)^2)(x+1), then determine its remainder.
Solution: Let Q(x) be the quotient, and the remainder be ax^2 + bx + c. Then, f(x)=((x-1)^2)(x+1)Q(x) + ax^2 + bx + c
If f(x) is divided by (x-1)^2, then its remainder is 3x+2. Thus, if ax^2 + bx + c is divided by (x-1)^2, then its remainder is 3x+2.
Therefore, f(x)=((x-1)^2)(x+1)Q(x)+a(x-1)^2 + 3x+2.
Since if f(x) is divided by x+1, then the remainder is 3, so from our new f(x) equation, we get
f(-1)=4a-3+2=3, thus a=1
Therefore, the final answer is (x-1)^2 + 3x+2=x^2 +x+3.

The part that I was having struggle with was the "...if ax^2 + bx + C is divided by (x-1)^2, then its remainder is 3x+2" part. I have no clue how in the world did the solution arrived with that part. Can anyone tell me how this part is done?

Thanks,

J.
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November 3rd, 2007, 08:41 PM   #2
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The solution states that the remainder is 3x + 2 when f(x) is divided by (x - 1)². It would seem, therefore, that the question had a typo in it, and should have given 3x + 2 instead of 3x + 1.
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November 4th, 2007, 07:22 AM   #3
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Yeah, I typed it wrong -- it should've been 3x+2. Thanks for letting me know.
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November 6th, 2007, 04:35 PM   #4
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Hmm... I still don't know how to arrive at the solution. Can anyone help me, if possible?
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