1. ## 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 $P=Q(x-1)+R$.

2. ## 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.

3. ## Re: Polynomial remainder problem

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 $...=Q(x+1)+\frac{5x-7}{x+1}$.

4. ## Re: Polynomial remainder problem

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.

5. ## Re: Polynomial remainder problem

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?

6. ## Re: Polynomial remainder problem

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.

7. ## Re: Polynomial remainder problem

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?

8. ## Re: Polynomial remainder problem

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.

9. ## Re: Polynomial remainder problem

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!

10. ## 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
Maths online