# Polynomial remainder problem

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• Oct 23rd 2012, 08:01 PM
PaulAdrienMauriceDirac
Polynomial remainder problem
Hello, this is a problem from Israel Gelfand's Algebra book about polynomials.

Problem 152. The polynomial $P$ gives a remainder of $5x-7$ when divided by $x^2-1$. Find the remainder when $P$ is divided by $x-1$.

My try: We know that $P=Q(x^2-1)+5x-7$ where Q is the quotient, this is the same as $P=Q(x+1)(x-1)+5x-7$, I see that all I have to do is to rewrite this equation to something like this $P=Q(x-1)+5x-7+r$, I tried dividing both sides by $x+1$ but I can't get the form of $P=Q(x-1)+R$.
Thanks in advance.
• Oct 23rd 2012, 08:05 PM
Prove It
Re: Polynomial remainder problem
\displaystyle \begin{align*} \frac{Q(x + 1)(x - 1) + 5x - 7}{x - 1} &= Q(x + 1) + \frac{5x - 7}{x - 1} \\ &= Q(x + 1) + \frac{5x - 5 - 2}{x + 1} \\ &= Q(x + 1) + \frac{5(x - 1)}{x - 1} - \frac{2}{x - 1} \\ &= Q(x + 1) + 5 - \frac{2}{x - 1} \end{align*}

So the remainder must be -2.
• Oct 23rd 2012, 08:17 PM
PaulAdrienMauriceDirac
Re: Polynomial remainder problem
Quote:

Originally Posted by Prove It
\displaystyle \begin{align*} \frac{Q(x + 1)(x - 1) + 5x - 7}{x - 1} &= Q(x + 1) + \frac{5x - 7}{x + 1} \\ &= Q(x + 1) + \frac{5x + 5 - 12}{x + 1} \\ &= Q(x + 1) + \frac{5(x + 1)}{x + 1} - \frac{12}{x + 1} \\ &= Q(x + 1) + 5 - \frac{12}{x + 1} \end{align*}

So the remainder must be -12.

Thank you for the reply Prove It, but I have few questions.
$\frac{P}{x-1}=\frac{Q(x+1)(x-1)+5x-7}{x-1}=Q(x+1)+\frac{5x-7}{x-1}$, isn't this correct? instead of $...=Q(x+1)+\frac{5x-7}{x+1}$.
• Oct 23rd 2012, 08:20 PM
Prove It
Re: Polynomial remainder problem
Quote:

Originally Posted by PaulAdrienMauriceDirac
Thank you for the reply Prove It, but I have few questions.
$\frac{P}{x-1}=\frac{Q(x+1)(x-1)+5x-7}{x-1}=Q(x+1)+\frac{5x-7}{x-1}$, isn't this correct? instead of $...=Q(x+1)+\frac{5x-7}{x+1}$.

Oops that's a typo. I'll fix it.
• Oct 23rd 2012, 08:28 PM
PaulAdrienMauriceDirac
Re: Polynomial remainder problem
Quote:

Originally Posted by Prove It
Oops that's a typo. I'll fix it.

It's ok, so should it be something like this?

\displaystyle \begin{align*}\frac{P}{x-1}=\frac{Q(x+1)(x-1)+5x-7}{x-1}=Q(x+1)+\frac{5x-7}{x-1}=Q(x+1)+\frac{5x-5-2}{x-1}=Q(x+1)+\frac{5(x-1)}{x-1}-\frac{2}{x-1}=Q(x+1)+5-\frac{2}{x-1}\end{align*}
and the remainder is -2? I don't understand this, can you explain me why is remainder -2 and not $5-\frac{2}{x-1}$? Is it because we divided $P$ by $x-1$ in the left side?
• Oct 23rd 2012, 08:30 PM
Prove It
Re: Polynomial remainder problem
Quote:

Originally Posted by PaulAdrienMauriceDirac
It's ok, so should it be something like this?

\displaystyle \begin{align*}\frac{P}{x-1}=\frac{Q(x+1)(x-1)+5x-7}{x-1}=Q(x+1)+\frac{5x-7}{x-1}=Q(x+1)+\frac{5x-5-2}{x-1}=Q(x+1)+\frac{5(x-1)}{x-1}-\frac{2}{x-1}=Q(x+1)+5-\frac{2}{x-1}\end{align*}
and the remainder is -2? I don't understand this, can you explain me why is remainder -2 and not $5-\frac{2}{x-1}$? Is it because we divided $P$ by $x-1$ in the left side?

The remainder is the numerator of the NON-WHOLE part. 5 is whole.
• Oct 23rd 2012, 08:46 PM
PaulAdrienMauriceDirac
Re: Polynomial remainder problem
Quote:

Originally Posted by Prove It
The remainder is the numerator of the NON-WHOLE part. 5 is whole.

Hm, can you tell me how did you know that, is that a rule in this type of situations, that remainder is the numerator of the non-whole part?
• Oct 23rd 2012, 10:02 PM
Prove It
Re: Polynomial remainder problem
Quote:

Originally Posted by PaulAdrienMauriceDirac
Hm, can you tell me how did you know that, is that a rule in this type of situations, that remainder is the numerator of the non-whole part?

It's a basic rule of writing improper fractions as mixed numbers.
• Oct 30th 2012, 11:40 PM
PaulAdrienMauriceDirac
Re: Polynomial remainder problem
Quote:

Originally Posted by Prove It
It's a basic rule of writing improper fractions as mixed numbers.

Yes I know that rule, I was just confused since I learned that $P=Q(Divisor)+R$, I was confused about which one was quotient and which one was remainder, now I understand, thank you!
• Oct 31st 2012, 09:51 PM
Salahuddin559
Re: Polynomial remainder problem
Remember that remainder is always less than the divisor, this is a little difficult to formalize in case of polynomials rather than in case of numbers. I think, a remainder when divided with a linear polynomial would be a constant, the acceptable remainder when you divide using a quadratic (like x^2 - 1) would be a linear polynomial (5x - 7) or a constant. Keep this in mind, and you will be able to factorize out correctly.

Salahuddin
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