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?

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- Aug 8th 2011, 10:57 AMbenny92000Remainder 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? - Aug 8th 2011, 11:17 AMSironRe: Remainder with polynomial division
Because that's the way you find a remainder.

For example if you want to solve the equation $\displaystyle f(x)=3x^4+7x^3+8x^2-2x-3=0$, in first instance you're going to search for a number $\displaystyle x_1$ wherefore $\displaystyle f(x_1)=0$, that means if you divide $\displaystyle f(x)$ by $\displaystyle (x-x_1)$ you'll obtain remainder 0, in this case if you enter -1 you see $\displaystyle f(x)=3$ so the remainder is 3. To have a closer look you can make the Horner scheme. - Aug 8th 2011, 11:48 AMAlso sprach ZarathustraRe: Remainder with polynomial division
- Aug 8th 2011, 11:57 AMbenny92000Re: 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? - Aug 8th 2011, 12:00 PMSironRe: Remainder with polynomial division
That was just an example (another case), and because $\displaystyle x_1$ is a zero therefore if you divive $\displaystyle f(x)$ by $\displaystyle (x-x_1)$ you'll obtain remainder 0 (this is a theoretical look to it).

- Aug 8th 2011, 12:49 PMbenny92000Re: Remainder with polynomial division
Oh... Ok. That seems to make sense now. Thanks!

- Aug 8th 2011, 12:52 PMSironRe: 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.