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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 (x1)^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 ((x1)^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)=((x1)^2)(x+1)Q(x) + ax^2 + bx + c If f(x) is divided by (x1)^2, then its remainder is 3x+2. Thus, if ax^2 + bx + c is divided by (x1)^2, then its remainder is 3x+2. Therefore, f(x)=((x1)^2)(x+1)Q(x)+a(x1)^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)=4a3+2=3, thus a=1 Therefore, the final answer is (x1)^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 (x1)^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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