Question

Partial fraction decomposition is going to be the death of me....
Here's another:
x^2+4x+4/x^3-x^2+2x-2

Answer 1

factorize to solve.

Answer 2

Nah, you know you secretly love partial fractions. If you go far in math you'll love them when you integrate, I swear.

As Saifoo states, you need to factor. Since I saw you got your other problem like this, you should be fine. The creators of the problem are getting fancy by making you factor a cubic equation. The secret for cubics is first look at the coefficients of the x^2 and the x... since they are -1 and 2 and they happen to multiply to give you -2 you are in some serious luck.
\[x^3-x^2+2x-2\] now becomes\[(x^2+2)(x-1)\]
Do as you did before and you'll get it.

Answer 3

(Ax+B)/(x^2+2) + C/(x-1)

Answer 4

Ok, so here is what I got but I don't think it is correct:
x^2+4x+4/(x^2+2)(x-1)
(Ax+B/x^2+2)+C/x-1
x^2+4x+4=(Ax+B)(x-1)+C(x^2+2)
x^2+4x+4=Ax^2-Ax+Bx-B+Cx^2+2C
x^2+4x+4=Ax^2+Cx^2-Ax+Bx-B+2C
Therefore:
A+C=1
-A+B=4
-B+2C=4
Solved:
A=-2/3
B=10/3
C=5/3
Therefore:
(-2/3x+10/3)/x^2+2 + (5/3)/x-1
Or:
-2x+10/2x^2+4 + 5/3x-1

Please help!!!!

Answer 5

(Ax+B)(x-1)+C(x^2+2)=x^2+4x+4
Ax^2-Ax+Bx-B+Cx^2+2C=x^2+4x+4
(A+C)x^2+x(-A+B)+(-B+2C)=1x^2+4x+4
A+C=1 => A=1-C
-A+B=4 => -(1-C)+B=4 (by substitution)=> C+B=5 => B=5-C
-B+2C=4 => -(5-C)+2C=4 (bu substition) => 3C=9 => C=3
B=5-3=2
A=1-3=-2

Answer 6

so i got
(-2x+2)/(x^2+2)+3/(x-1)

Answer 7

we can check this by combining the fractions

Answer 8

(-2x+2)(x-1)+3(x^2+2)
-----------------------
(x^2+2)(x-1)


=


-2x^2+2x+2x-2+3x^2+6
--------------------------
(x^2+2)(x-1)

=
x^2+4x+4
---------------
(x^2+2)(x-1)

our check verifies we have it correct

Answer 9

You had up to the system of equations correct. I think substitution is a ugly method for this, try addition method.

(1) A+C=1
(2) -A+B=4
(3) -B+2C=4

Add (1) and (2) together and the A value will cancel out.

B+C=5

Now add this new equation to (3) You should have the B value cancel out

3C=9
C=3
Now plug it in backwards to get
A=-2
B= 2

Answer 10

sub works and i think its pretty lol

Answer 11

You are both wonderful! I can't believe I had it right up until the most basic substitution part! Thanks!

Answer 12

Lol, Substitution does work, but sometimes is makes things not look so pretty. I only use it for systems with 2 equations. When you have 3 equations it is more efficient to use Addition. :P

Shilohk, you have a good understanding of this stuff, just double check things plug plugging them in. Partial fractions usually do not have nasty fractions like you got and it was good you thought you did something wrong from it. When it comes to the test you'll be better prepared.