- Thread starter
- #1

- Thread starter gobindo
- Start date

- Thread starter
- #1

- Admin
- #2

By the division algorithm, we may state:

\(\displaystyle P(x)=(x-1)(x-2)Q(x)+R(x)\)

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

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

And from the remainder theorem we know:

\(\displaystyle P(1)=1\)

\(\displaystyle P(2)=3\)

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:

\(\displaystyle P(1)=3\)

\(\displaystyle P(2)=1\)

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?

- Aug 30, 2012

- 1,197

You are going to win the "most psychic member" award this year... Good call.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 $-2x+5$.

-Dan

- Admin
- #4

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

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

Subtract the second equation from the first:

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

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

\(\displaystyle Q(x)\equiv Q_2(x)-Q_1(x)\) ?

- Thread starter
- #5

thank you mark for the quick response ,also you guessed it right that it was misquoted as =.well done.

By the division algorithm, we may state:

\(\displaystyle P(x)=(x-1)(x-2)Q(x)+R(x)\)

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

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

And from the remainder theorem we know:

\(\displaystyle P(1)=1\)

\(\displaystyle P(2)=3\)

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:

\(\displaystyle P(1)=3\)

\(\displaystyle P(2)=1\)

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?