Polynomial remainder problem

**PaulAdrienMauriceDirac** · October 23rd 2012, 07:01 PM

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.

**Prove It** · October 23rd 2012, 07:05 PM

$\displaystyle \begin{align*} \frac{Q(x + 1)(x - 1) + 5x - 7}{x - 1} &= Q(x + 1) + \frac{5x - 7}{x - 1} \\ &= Q(x + 1) + \frac{5x - 5 - 2}{x + 1} \\ &= Q(x + 1) + \frac{5(x - 1)}{x - 1} - \frac{2}{x - 1} \\ &= Q(x + 1) + 5 - \frac{2}{x - 1} \end{align*}$

So the remainder must be -2.

**PaulAdrienMauriceDirac** · October 23rd 2012, 07:17 PM

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

**Prove It** · October 23rd 2012, 07:20 PM

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.

**PaulAdrienMauriceDirac** · October 23rd 2012, 07:28 PM

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?

**Prove It** · October 23rd 2012, 07:30 PM

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.

**PaulAdrienMauriceDirac** · October 23rd 2012, 07:46 PM

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?

**Prove It** · October 23rd 2012, 09:02 PM

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.

**PaulAdrienMauriceDirac** · October 30th 2012, 10:40 PM

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!

**Salahuddin559** · October 31st 2012, 08:51 PM

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

Thread: Polynomial remainder problem

LinkBack

Thread Tools

Display

Polynomial remainder problem

Re: Polynomial remainder problem

Re: Polynomial remainder problem

Re: Polynomial remainder problem

Re: Polynomial remainder problem

Re: Polynomial remainder problem

Re: Polynomial remainder problem

Re: Polynomial remainder problem

Re: Polynomial remainder problem

Re: Polynomial remainder problem

Similar Math Help Forum Discussions

Taylor Polynomial and Remainder

Remainder of a polynomial

polynomial division, remainder

Remainder, equation of a polynomial

Taylor Polynomial & Remainder

Search Tags