# Thread: long division

1. ## long division

hi, I have no idea how the solution turned this improper fraction into a proper one. Question is solve this using partial fractions: $\displaystyle \frac{x^3+2x^2+3x+1}{x^2(x+1)}=1+\frac{x^2+3x+1}{x ^2(x+1)}$ I dont know how they got the second fraction through long division could someone show me?

2. Originally Posted by oxrigby
hi, I have no idea how the solution turned this improper fraction into a proper one. Question is solve this using partial fractions: $\displaystyle \frac{x^3+2x^2+3x+1}{x^2(x+1)}=1+\frac{x^2+3x+1}{x ^2(x+1)}$ I dont know how they got the second fraction through long division could someone show me?

$\displaystyle x^3 + 2x^2 + 3x + 1$

$\displaystyle = x^3 + x^2 + x^2 + 3x + 1$

and note this $\displaystyle x^3 + x^2 = x^2(x+1)$

$\displaystyle x^3 + x^2 + x^2 + 3x + 1$

$\displaystyle x^2(x+1) + x^2 + 3x+1$

3. Originally Posted by oxrigby
hi, I have no idea how the solution turned this improper fraction into a proper one. Question is solve this using partial fractions: $\displaystyle \frac{x^3+2x^2+3x+1}{x^2(x+1)}=1+\frac{x^2+3x+1}{x ^2(x+1)}$ I dont know how they got the second fraction through long division could someone show me?
I'm not good enough at LaTex to draw this well but:

$\displaystyle x^2(x+1)= x^3+ x^2+ 0x+ 0$ so you are dividing $\displaystyle x^3+ 2x^2+ 3x+ 1$. As a "trial division", $\displaystyle x^3$ divides into $\displaystyle x^3+ 2x^2+ 3x+ 1$ once. Now subract: x^3+ 2x^2+ 3x+ 1[/tex] minus $\displaystyle x^3+ x^2$ is $\displaystyle x^2+ 3x+ 1$. Since that has degree lower than 3, $\displaystyle x^3+ 2x^2+ 3x+ 1= 1(x^3+ 2x^2)+ (x^2+ 3x+ 1)$ or $\displaystyle \frac{x^3+ 2x^2+ 3x+ 1}{x^3+ 2x^2}= 1+ \frac{x^2+ 3x+ 1}{x^3+ 2x^2}$.

4. I don't know latex well enough to give visuals

$\displaystyle \frac {x^3 + 2x^2 + 3x + 1}{x^2(x + 1)} = \frac {x^3 + 2x^2 + 3x + 1}{x^3 + x^2}$

First thing we do in long division, how many times does the first term in the denominator go into the first term in the numerator? That is, how many times does x^3 go into x^3? The answer is 1, so bring the 1 up and multiply the denom by that amount and subtract it from the numerator.

$\displaystyle (x^3 + 2x^2 + 3x + 1) - (x^3 + x^2) = x^2 + 3x + 1$

They got lazy and stopped there, but this would be your remainder $\displaystyle (x^2 + 3x + 1) / (x^3 + x^2)$

So that was $\displaystyle 1 + (x^2 + 3x + 1) / (x^3 + x^2)$

5. aidan, you need "[ math ]" "[ \math ]" tags (without the spaces).

6. $\displaystyle x^3+x^2|x^3+2x^2+3x+1 -x^3+x^2 x^2+3x$
gives me 1+1 on the top so far ,,cant possibly be correct?

7. $\displaystyle \frac{x^3+2x^2+3x+1}{x^2(x+1)}=1+\frac{x^2+3x+1}{x ^2(x+1)}$

Notice the degree of the top in the last fraction is lower than the degree of the bottom. We won't get another 1.

8. does long division not work then?''.

9. ---------------------
x^3 + x^2 | x^3 + 2x^2 + 3x + 1
How many times does x^3 go into x^3?

1
---------------------
x^3 + x^2 | x^3 + 2x^2 + 3x + 1
Now multiply that 1 by (x^3 + x^2) and subtract
1
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x^3 + x^2 | x^3 + 2x^2 + 3x + 1
-(x^3 + x^2)
= x^2 + 3x + 1
Seeing as x^3 doesn't go nicely into this next term, we just leave it as the remainder.

So $\displaystyle 1 + \frac{x^2 + 3x + 1}{x^3 + x^2}$

We could, however, continue to break it up if you'd like:

$\displaystyle 1 + \frac 1{x+1} + \frac 2x + \frac 1{x^2}$