# Remainder with polynomial division

• August 8th 2011, 11:57 AM
benny92000
Remainder with polynomial division
The question: The remainder obtained when 3x^4 + 7x^3 + 8x^2 -2x -3 is divided by x + 1 is?

The answer is 3. My book says to plug in -1 for x (which results with 3). Why should I do this?
• August 8th 2011, 12:17 PM
Siron
Re: Remainder with polynomial division
Because that's the way you find a remainder.
For example if you want to solve the equation $f(x)=3x^4+7x^3+8x^2-2x-3=0$, in first instance you're going to search for a number $x_1$ wherefore $f(x_1)=0$, that means if you divide $f(x)$ by $(x-x_1)$ you'll obtain remainder 0, in this case if you enter -1 you see $f(x)=3$ so the remainder is 3. To have a closer look you can make the Horner scheme.
• August 8th 2011, 12:48 PM
Also sprach Zarathustra
Re: Remainder with polynomial division
Quote:

Originally Posted by benny92000
The question: The remainder obtained when 3x^4 + 7x^3 + 8x^2 -2x -3 is divided by x + 1 is?

The answer is 3. My book says to plug in -1 for x (which results with 3). Why should I do this?

Polynomial long division - Wikipedia, the free encyclopedia
• August 8th 2011, 12:57 PM
benny92000
Re: Remainder with polynomial division
Hmm, thanks. Makes a little more sense. I guess I still need to work a few hands-on examples to make sense of why we do that. Let me ask a few questions to get a more thorough understanding:
1.) Why do we set the first equation equal to 0?
2.) Why does dividing by the the x-x1 give us remainder 0?
• August 8th 2011, 01:00 PM
Siron
Re: Remainder with polynomial division
That was just an example (another case), and because $x_1$ is a zero therefore if you divive $f(x)$ by $(x-x_1)$ you'll obtain remainder 0 (this is a theoretical look to it).
• August 8th 2011, 01:49 PM
benny92000
Re: Remainder with polynomial division
Oh... Ok. That seems to make sense now. Thanks!
• August 8th 2011, 01:52 PM
Siron
Re: Remainder with polynomial division
You're welcome!
Beside the 'polynomial long division' (see Also sprach Zarathrusta's post) you can also use Horner scheme to evaluate the remainder.