# Write the expression ... in the form ...

• Oct 16th 2012, 11:41 AM
TheCracker
Write the expression ... in the form ...
I kinda don't understand that assignement, can someone help me with it please? What do I need to do?

http://imageshack.us/a/img717/650/clipboard01jzd.jpg

• Oct 16th 2012, 12:35 PM
johnsomeone
Re: Write the expression ... in the form ...
$\text{If you divide 6 into 29, you get 4 with remainder 5, so } \frac{29}{6} = 4 + \frac{5}{6}$.

Polynomials behave the same way, only now you stop with a "remainder" when the degree is less than the degree of the denominator (as compared to integers, where you stop when the numerator is less than the demoninator.) But it's still the procedure of rewriting a fraction after doing the long division and stopping with a remainder. It's analogous to converting a simple fraction into a mixed fraction.

$\text{Example 1: }\frac{x^2-4x+1}{x+2} = (x-6) + \frac{13}{x+2}. \text{ Note the numerator is a degree 0 polynomial in x.}$

$\text{Check: }(x-6) + \frac{13}{x+2} = \frac{(x-6)(x+2)}{x+2} + \frac{13}{x+2} = \frac{(x^2 - 4x - 12) + (13)}{x+2}$

$= \frac{x^2-4x+1}{x+2}.$

$\text{Example 2: }\frac{x^2-4x+1}{x^2+2} = 1 + \frac{-4x-1}{x^2+2}. \text{ Note the numerator is a degree 1 polynomial in x.}$

$\text{Check: } 1 + \frac{-4x-1}{x^2+2} = \frac{x^2+2}{x^2+2} + \frac{-4x-1}{x^2+2} = \frac{(x^2+2) + (-4x-1)}{x^2+2}$

$= \frac{x^2-4x+1}{x^2+2}.$

So what you need to do is polynomial long division. Do you know how to do that?
• Oct 16th 2012, 12:52 PM
Soroban
Re: Write the expression ... in the form ...
Hello, TheCracker!

What a pendantic way to write the problem!
No wonder you "kinda don't understand" it.

Quote:

$\text{Write the expression }\,\frac{x^6 + 6x^3 + 4}{x^2+2x+5}\,\text{ in the form }\,G(x) + \frac{R(x)}{q(x)}$

$\text{where }q(x) \,=\,x^2+2x+5\,\text{ is the denominator of the given expression,}$

$\text{and }G(x)\text{ and }R(x)\text{ are polonimials with }\text{deg(}R)\,<\,\text{deg}(G).$

They are asking us to perform some long division.

We are to find the quotient $G(x)$
. . and write the remainder in the form of a fraction: . $\frac{R(x)}{x^2+2x+5}$

$\begin{array}{cccccccccccccccc} &&&&&& x^4 &-& 2x^3 &-& x^2 &+& 18x &-& 31 \\ && --&--&--&--&--&--&--&--&--&--&--&--&-- \\ x^2+2x+5 & | & x^6 &&&&& +& 6x^3 &&&&& + & 4 \\ && x^6 &+& 2x^5 &+& 5x^4 \\ && --&--&--&--&-- \\ &&&& -2x^5 &-& 5x^4 &+& 6x^3 \\ &&&& -2x^5 &-& 4x^4 &-& 10x^3 \\ &&&& --&--&--&--&-- \\ &&&&& -& x^4 &+& 16x^3 \\ &&&&& -& x^4 &-& 2x^3 &-& 5x^2 \\ &&&&& --&--&--&--&--&-- \\ &&&&&&&& 18x^3 &+& 5x^2 \\ &&&&&&&& 18x^3 &+& 36x^2 &+& 90x \\ &&&&&&&& --&--&--&--&-- \\ &&&&&&&&& -& 31x^2 &-& 90x &+& 4 \\ &&&&&&&&& -& 31x^2 &-& 62x &-& 155 \\ &&&&&&&&& --&--&--&--&--&-- \\ &&&&&&&&&&& -& 28x &+& 159 \end{array}$

Therefore: . $\frac{x^6 + 6x^3 + 4}{x^2+2x+5} \;\;=\;\;x^4 - 2x^3 - x^2 + 18x - 31 + \frac{\text{-}28x + 159}{x^2+2x+5}$

• Oct 16th 2012, 01:57 PM
TheCracker
Re: Write the expression ... in the form ...
Okay, now I understand it, but how do I enter it into the available fields?
Do I enter it like x^4-2x^3-x^2 and so forth? I'm a little bit confused, because normally you have like an area you can write in when you have something to the power of something, but in such empty fields they always ask for specific real numbers :/
• Oct 17th 2012, 03:38 AM
TheCracker
Re: Write the expression ... in the form ...
Any ideas? :/
• Oct 17th 2012, 04:10 AM
earboth
Re: Write the expression ... in the form ...
Quote:

Originally Posted by TheCracker
Okay, now I understand it, <-- are you sure?
but how do I enter it into the available fields?
Do I enter it like x^4-2x^3-x^2 and so forth? I'm a little bit confused, because normally you have like an area you can write in when you have something to the power of something, but in such empty fields they always ask for specific real numbers :/

Quote:

Originally Posted by TheCracker
Any ideas? :/

$\frac{x^6 + 6x^3 + 4}{x^2+2x+5} \;\;=\;\;\underbrace{x^4 - 2x^3 - x^2 + 18x - 31}_{G(x)} + \frac{\overbrace{\text{-}28x + 159}^{R(x)}}{x^2+2x+5}$
• Oct 17th 2012, 06:03 AM
TheCracker
Re: Write the expression ... in the form ...
I wasn't sure if it would work with the fields provided, because as I said, normally when you have x to the power of something, the input-field looks different

Nevermind though, because it worked

Thank you