+108 Could anybody help my decompose this fraction? Thanks!
(x^3-2)/(x^2+x)
When you have a hard time, just expand the equation:
(x^3-2)/ (x^2-2) : (X)(X)(X)-2 / (x)(x)+X
now proceed to eliminate which gives: (X)-2 / X
X/X= 1 so they cancel out
-2/1 = -2
:)
Simplify the following:
(x^3 - 2)/(x^2 + x)
There isn't much you can do with it !!
Factor common terms out of x^2 + x.
Factor x out of x^2 + x:
(x^3 - 2)/(x (x + 1)), or you can write it like this:x^3/(x^2 + x) - 2/(x^2 + x)
+118687 I am not sure
All I can think to do is division which gave me
\(\frac{x^3-2}{x^2+x}\\ =x-1+\frac{x-2}{x^2+2x} \qquad \text{(This has been edited)}\)
I will leave this to another mathematician.
Continued from earlier
\(\frac{x^3-2}{x^2+x}\\ =x-1+\frac{x-2}{x^2+2x}\\ =x-1+\frac{x-2}{x(x+2)}\\\)
consider just the fraction part
\(\frac{x-2}{x(x+2)}=\frac{A}{x}+\frac{B}{x+2} \qquad \text{for some real A and B}\\ A(x+2)+Bx\equiv x-2\\ Ax+2A+Bx\equiv x-2\\ (A+B)x+2A\equiv x-2\\ so\\ 2A=-2\\ A=-1\\ A+B=1\\ -1+B=1\\ B=2\\ \frac{x-2}{x(x+2)}=\frac{-1}{x}+\frac{2}{x+2} \)
So
\(\frac{x^3-2}{x^2+x} \\ =x-1+\frac{x-2}{x^2+2x} \\ =x-1+\frac{-1}{x}+\frac{2}{x+2}\\ =x-1-\frac{1}{x}+\frac{2}{x+2}\\\)
I think that is right now. 
What do you think Mathbum and guest?
+118687 LOL this must not be my question! it is jinxed 
Oh well doesn't matter, the technique is the same.
Mathbum can copy the technique and get the original answer correct :)
+108 At the consider just the fraction part Melody, why would you have x-2 in the numerator on the left side?
+118687 Becasue I did the algebraic division first (which I did not display) and x-2 was the remiander.
\((x^2+x) \text{ divided into } (x^3-2) \;\;\;goes \;\;\; (x-1) \text{ times with a remainder of } (x-2)\)
\(so\\ \frac{x^3-2}{x^2+x}=x-1+\frac{x-2}{x^2+x} \)
Maybe this will help
\(7 \div 2=3\;\;remainder 1\\ so\\ \frac{7}{2}=3+\frac{1}{2}\)
I hope I have not made any typing errors this time. 
Do you know how to do the algebraic division?
+82 
So basically you just have to factor out the common factors on both the denominator or numberstor then cancel them out.
hope this helps!
+129899 (x^3 - 2) / (x^2 + x)
Partial faction decomposition can only be done whenever the degree of the polynomial in the numerator is < the degree of the polynomial in the denominator
Performing polynomial division, we have
x - 1
x^2 + x [ x^2 + 0x^2 + 0x - 2 ]
x^3 + x^2
___________________
-x^2 + 0x
-x^2 - 1x
_________
x - 2
The result is ( x - 1) + ( x - 2) / ( x^2 + x)
We can decompose the remainder as
( x - 2) / [ x ( x + 1) = A / x + B / ( x + 1)
Multiply through by x ( x + 1)
x - 2 = A(x + 1) + Bx simplify
x - 2 = (A + B)x + A equate coefficients
A + B = 1
A = -2
Which means that B = 3
So...the complete decomposition is
(x -1 ) - [ 2 / x ] + [ 3 / ( x + 1) ]
