MHB Prove Polynomial Remainder: -2x+5 When Divided by (x-1)(x-2)

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The discussion revolves around proving that a polynomial P(x) leaves a remainder of -2x + 5 when divided by (x-1)(x-2). Participants agree that the problem may have been misquoted, suggesting that the remainders for P(1) and P(2) were reversed. By applying the division algorithm and the remainder theorem, they establish a linear function for the remainder. A system of equations is proposed to solve for the parameters of the linear remainder. The consensus is that the correct remainder is indeed -2x + 5.
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A polynomial P(x) when divided by(x-1) leaves a remainder 1 and when divided by (x-2) leaves a remainder of3. prove that when divided by(x-1(x-2) it leaves a remainder -2x=5.
thank you.
 
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Re: I'm having trouble with this question:

By the division algorithm, we may state:

$$P(x)=(x-1)(x-2)Q(x)+R(x)$$

We know the remainder must be a linear function (right?), and so we may state:

(1) $$P(x)=(x-1)(x-2)Q(x)+ax+b$$

And from the remainder theorem we know:

$$P(1)=1$$

$$P(2)=3$$

I suspect the problem has been misquoted, since the given remainder is not correct, whether the "=" should be "+" or "-". I am assuming then that the constant remainders have been reversed, and the linear remainder is in fact $-2x+5$. So, we have instead:

$$P(1)=3$$

$$P(2)=1$$

Using the two equations above and (1), we may get a 2 X 2 linear system in the parameters $a$ and $b$, which will have a unique solution. Can you put all of this together?
 
Re: I'm having trouble with this question:

MarkFL said:
I suspect the problem has been misquoted, since the given remainder is not correct, whether the "=" should be "+" or "-". I am assuming then that the constant remainders have been reversed, and the linear remainder is in fact $-2x+5$.
You are going to win the "most psychic member" award this year... Good call.

-Dan
 
Here's an alternate approach:

$$\frac{P(x)}{x-2}=Q_2(x)+\frac{1}{x-2}$$

$$\frac{P(x)}{x-1}=Q_1(x)+\frac{3}{x-1}$$

Subtract the second equation from the first:

$$P(x)\left(\frac{1}{x-2}-\frac{1}{x-1} \right)=\left(Q_2(x)-Q_1(x) \right)+\left(\frac{1}{x-2}-\frac{3}{x-1} \right)$$

What do you find upon simplification, and using the definition:

$$Q(x)\equiv Q_2(x)-Q_1(x)$$ ?
 
Re: I'm having trouble with this question:

thank you mark for the quick response ,also you guessed it right that it was misquoted as =.well done.
MarkFL said:
By the division algorithm, we may state:

$$P(x)=(x-1)(x-2)Q(x)+R(x)$$

We know the remainder must be a linear function (right?), and so we may state:

(1) $$P(x)=(x-1)(x-2)Q(x)+ax+b$$

And from the remainder theorem we know:

$$P(1)=1$$

$$P(2)=3$$

I suspect the problem has been misquoted, since the given remainder is not correct, whether the "=" should be "+" or "-". I am assuming then that the constant remainders have been reversed, and the linear remainder is in fact $-2x+5$. So, we have instead:

$$P(1)=3$$

$$P(2)=1$$

Using the two equations above and (1), we may get a 2 X 2 linear system in the parameters $a$ and $b$, which will have a unique solution. Can you put all of this together?
 
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