Question

Note: This is NOT a question. This is a tutorial.

how to divide a polynomial by a binomial? For example you are given \(\large \frac{x^3 + 2x^2 - 3x + 4}{x - 5}\)

See comment below to see how!

Answer 1

The first step is to look at the degrees. The highest exponent of the numerator must be GREATER than the highest exponent in the denominator so we can divide them. In the example, the highest exponent of the numerator is 3 (from \(x^3\), and the highest exponent of the denominator is 1 (from \(x^1\) so we can divide them.

_________________________
\(x - 5\) | \(x^3 + 2x^2 - 3x + 4\)

The first step is to divide the first term of the dividend (which is \(x^3\)) by the first term of the divisor (which is \(x\)) so we have \(\frac{x^{\cancel{3}}}{\cancel{x}} = x^2\)

\(x^2\)
_________________________
\(x - 5\) | \(x^3 + 2x^2 - 3x + 4\)

Now, we multiply the quotient (which is \(x^2\)) to the first and second term of the divisor, and we'll put it beneath the numbers of the dividend, just like in ordinary division.

\(x^2\)
_________________________
\(x - 5\) | \(x^3 + 2x^2 - 3x + 4\)
\(x^3 - 5x^2\)

Now, we subtract. Remember to change the signs of the subtrahend when subtracting.

\(x^2\)
_________________________
\(x - 5\) | \(x^3 + 2x^2 - 3x + 4\)
\(x^3 - 5x^2\)
--------------
\(0 + 7x^2\)

Now, we divide the difference by the first term of the divisor again.


\(x^2 + 7x\)
_________________________
\(x - 5\) | \(x^3 + 2x^2 - 3x + 4\)
\(x^3 - 5x^2\)
--------------
\(0 + 7x^2\)

Then, we multiply the quotient by the first and second term of the divisor again. We will also bring down the -3x from the dividend.

\(x^2 + 7x\)
_________________________
\(x - 5\) | \(x^3 + 2x^2 - 3x + 4\)
\(x^3 - 5x^2\)
--------------
\(0 + 7x^2 - 3x\)
\(7x^2 - 35x\)

We subtract again.


\(x^2 + 7x\)
_________________________
\(x - 5\) | \(x^3 + 2x^2 - 3x + 4\)
\(x^3 - 5x^2\)
--------------
\(0 + 7x^2 - 3x\)
\(7x^2 - 35x\)
------------
\(0 + 32x\)

Then we divide it by the divisor again.

\(x^2 + 7x + 32\)
_________________________
\(x - 5\) | \(x^3 + 2x^2 - 3x + 4\)
\(x^3 - 5x^2\)
--------------
\(0 + 7x^2 - 3x\)
\(7x^2 - 35x\)
------------
\(0 + 32x\)

Multiply the quotient by the first and second term of the divisor and bring down +4...


\(x^2 + 7x + 32\)
_________________________
\(x - 5\) | \(x^3 + 2x^2 - 3x + 4\)
\(x^3 - 5x^2\)
--------------
\(0 + 7x^2 - 3x\)
\(7x^2 - 35x\)
------------
\(0 + 32x + 4\)
\(32x - 160\)

Subtract.


\(x^2 + 7x + 32\)
_________________________
\(x - 5\) | \(x^3 + 2x^2 - 3x + 4\)
\(x^3 - 5x^2\)
--------------
\(0 + 7x^2 - 3x\)
\(7x^2 - 35x\)
------------
\(0 + 32x + 4\)
\(32x - 160\)
----------
0 + 164

Now, we cannot divide it any longer. This becomes the remainder. we will now add the final quotient (which is \(x^2 + 7x + 32\)) to the remainder-over-divisor. The remainder is 164, and the divisor is x - 5. So, we'll add \(\frac{164}{x-5}\) to the final quotient. And so, our final answer is

\(\large x^2 + 7x + 32 + \frac{164}{x-5}\)

Answer 2

sorry if they arent aligned very much. It was aligned when i typed it in notepad :/

Answer 3

damn nice formatting

Answer 4

haha thanks :)

Answer 5

yeah, iggy... that's not even latex...

Answer 6

The real tragedy is that the people who need this won't read it. :\

Answer 7

@dpaInc it's latex lol =)) doesnt it look like one? @badreferences im sure someday this will be noticed :) maybe if there would be a group where people could post their own tutorials...

Answer 8

nope, this thread will get buried

Answer 9

hmm that's encouraging lol =)) well it's a good time killer. If anyone wanna see it they could just check it in my profile. They're all there...

Answer 10

I honestly think that this tutorial thing is awesome, it just might get somewhere :D

Answer 11

haha thanks for that goodie ^_^ i hope so too

Answer 12

maybe i should do some on harder topics..hmmm...maybe thatll increase the credibility and publicity of these tutorials

Answer 13

don't worry I will keep copy/pasting your work and citing you

Answer 14

oh yay thanks :D

Answer 15

Ohhh, do calculus ones, i wanna learn how to do calculus before i learn it. So i seem smart :D

Answer 16

@GOODMAN Calculus is a huge topic. But I guess I could start you on basic rules? Await my return. :P

Answer 17

i started some of the few basic rules @badreferences maybe you can add some with a touch of your methods as well? that would be helpful :D

Answer 18

Thanks :).