# Polynomial Long Division w Integrals

#### shamieh

##### Active member
Sep 13, 2013
539
I do not understand how I would do this with long division since there is only 2 terms. I can't remember the trick. Here is what I have so far.
$$\displaystyle \int \frac{3x^2 - 2}{x^2 - 2x - 8} dx$$

so I got $$\displaystyle \int 3 + \frac{x^2 - 2}{(x - 4)(x + 2)}$$

I'm not sure if that's right? I just factored it out instead of long division. I Don't recall what "synthetic division" is either.

So I have $$\displaystyle (x - 4)(x + 2) = A(x+2) + B(x-4)$$

Am I on the right track or no?

#### Ackbach

##### Indicium Physicus
Staff member
Jan 26, 2012
4,197
Synthetic division is possible, but not usually recommended when the denominator is quadratic or higher. To do polynomial long division, just make sure that the dividend and the divisor have all the descending powers of $x$. If there are any missing powers, add it in as follows:
$$\frac{3x^2-2}{x^2 -2x-8}= \frac{3x^2+0x-2}{x^2-2x-8}.$$
Now do the usual long division algorithm.

Last edited:

#### Deveno

##### Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
I do not understand how I would do this with long division since there is only 2 terms. I can't remember the trick. Here is what I have so far.
$$\displaystyle \int \frac{3x^2 - 2}{x^2 - 2x - 8} dx$$

so I got $$\displaystyle \int 3 + \frac{x^2 - 2}{(x - 4)(x + 2)}$$

I'm not sure if that's right? I just factored it out instead of long division. I Don't recall what "synthetic division" is either.

So I have $$\displaystyle (x - 4)(x + 2) = A(x+2) + B(x-4)$$

Am I on the right track or no?

I would re-write:

$3x^2 - 2 = 3(x^2 - 2x - 8) + 6x + 22$

$$\displaystyle \int \frac{3x^2 - 2}{x^2 - 2x - 8}dx = \int 3\ dx + \int\frac{6x + 22}{x^2 - 2x - 8}dx$$

Next, re-write $6x + 22$ as $6(x + 2) + 10$

so that:

$$\displaystyle \int\frac{6x + 22}{x^2 - 2x - 8}dx = \int \frac{6}{x - 4}dx + \int\frac{10}{x^2 - 2x - 8}dx$$

Now you're in a position to use "partial fraction decomposition", by solving:

$A(x - 4) + B(x + 2) = 10$ leading to:

$A + B = 0$
$2B - 4A = 10$

which should not be that hard.

#### shamieh

##### Active member
Sep 13, 2013
539
I would re-write:

$3x^2 - 2 = 3(x^2 - 2x - 8) + 6x + 22$
What is going on here Deveno? Your setting the numerator = to the denominator? Where are those extra terms coming from? Are you completing the square? I am lost, can you show me a little detail on this particular step just so I can understand.

#### MarkFL

$$\displaystyle 3x^2-2=3x^2+6x-6x-24+22=3\left(x^2-2x-8 \right)+6x+22$$
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