My Math Forum An Algebraic Long Division, yet the solution is very tricky!

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 November 2nd, 2007, 06:52 PM #1 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 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.
 November 3rd, 2007, 08:41 PM #2 Global Moderator   Joined: Dec 2006 Posts: 21,028 Thanks: 2259 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.
 November 4th, 2007, 07:22 AM #3 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Yeah, I typed it wrong -- it should've been 3x+2. Thanks for letting me know.
 November 6th, 2007, 04:35 PM #4 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Hmm... I still don't know how to arrive at the solution. Can anyone help me, if possible?

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