Author Topic: Polynomials - long division and remainder theorum (Read 1247 times) Tweet Share

fossil · « **on:** February 26, 2010, 06:07:19 pm »

Hi
Stuck on a couple of questions from this topic, it'd be great if someone could help me out with these:

1. $x^4-9x^3+25x^2-8x-2$ divided by $x^2-2$

2. Let $P(x)=x^5-3x^4+2x^3-2x^2+3x+1$ . Given that P(x) can be written in the form $(x^2-1)Q(x)+ax+b$ where Q(x) is a polynomial and a and b are constants, hence or otherwise, find the remainder when P(x) is divided by $x^2-1$ .

Thanks

ZachCharge · « **Reply #1 on:** February 26, 2010, 08:32:07 pm »

Hmm, this will be a bit hard to show working out on Latex, but I will try;

   $x^2$
   __________________________
$x^2-2$ | $x^4-9x^3+25x^2-8x-2$
   $x^4+0-2x^2$

$x^2-9x$
   __________________________
$x^2-2$ | $x^4-9x^3+25x^2-8x-2$
   $x^4+0-2x^2$
   __________________________
   $-9x^3+27x^2-8x$
   $9x^3+0+18x$


   $x^2-9x+27$
   __________________________
$x^2-2$ | $x^4-9x^3+25x^2-8x-2$
   $x^4+0-2x^2$
   __________________________
   $-9x^3+27x^2-8x$
   $9x^3+0+18x$
   __________________________
   $27x^2-26x-2$
   $27x^2+0-54$
   __________________________
   $-26x+52$

$Q(x)=x^2-9x+27$
$R(x)=52-26x$

fossil · « **Reply #2 on:** February 26, 2010, 10:16:14 pm »

Thanks, that's really clear.
Explains both questions too, although I'm wondering how you'd work out the second question "hence"...

the.watchman · « **Reply #3 on:** February 26, 2010, 10:30:03 pm »

1) $\frac{x^4 - 9x^3 + 25x^2 - 8x - 2}{x^2 - 2} = \frac{x^2 (x^2 - 2) - 9x (x^2 - 2) + 27(x^2 - 2)}{x^2 - 2} - \frac{-26x+52}{x^2 - 2}$

$\frac{x^4 - 9x^3 + 25x^2 - 8x - 2}{x^2 - 2} = x^2 - 9x + 27 - \frac{-26x+52}{x^2 - 2}$

2) Find $\frac{P(x)}{x^2 - 1}$ and use this to find Q(x) and the remainder (ax+b)

ZachCharge · « **Reply #4 on:** February 26, 2010, 10:35:08 pm »

Err....what happened to my answer lol.

EDIT: Oh that is just evil, the topic name really needs to include something hinting to exam discussion lol.

the.watchman · « **Reply #5 on:** February 26, 2010, 10:38:18 pm »

Quote from: ZachCharge on February 26, 2010, 10:35:08 pm

Err....what happened to my answer lol.

EDIT: Oh that is just evil, the topic name really needs to include something hinting to exam discussion lol.

LOL this thread is repeated twice, I answered this one, and you answered the other one

kyzoo · « **Reply #6 on:** February 27, 2010, 09:01:48 am »

0.o....What the hell, that's the exact same question as this

http://vcenotes.com/forum/index.php/topic,22087.msg229365.html#msg229365

fossil · « **Reply #7 on:** February 27, 2010, 09:47:55 am »

Thanks for that - yeah sorry, I put it in the wrong topic and then it wouldn't let me delete it so I just copied it into the methods forum.

TrueTears · « **Reply #8 on:** February 27, 2010, 09:52:02 am »

Quote from: fossil on February 27, 2010, 09:47:55 am

Thanks for that - yeah sorry, I put it in the wrong topic and then it wouldn't let me delete it so I just copied it into the methods forum.

Moved and merged

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Author Topic: Polynomials - long division and remainder theorum (Read 1247 times) Tweet Share

fossil

Polynomials - long division and remainder theorum

ZachCharge

Re: Polynomials - long division and remainder theorum

fossil

Re: Polynomials - long division and remainder theorum

the.watchman

Re: Polynomials - long division and remainder theorum

ZachCharge

Re: Polynomials - long division and remainder theorum

the.watchman

Re: Polynomials - long division and remainder theorum

kyzoo

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fossil

Re: Polynomials - long division and remainder theorum

TrueTears

Re: Polynomials - long division and remainder theorum

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