Expressing this algebraic fraction as partial fraction

[MadmanMurray]

Homework Statement

x³ - x² - 5x / x² -3x +2

Homework Equations

The Attempt at a Solution

Heres what I got
5/x-1 + 6/x-2
but if I multiply those 2 fractions together I do not get a cubic equation on top of the line. I learned how to do these questions with linear equations on top I'm lost here though. Is there some trick that I don't know about?

 

[Delphi51]

I don't see how the 5/x-1 + 6/x-2 is related to the question. Could you describe what you did? Certainly this is not equal to the original expression. For example, when x=0, the given expression evaluates to zero while your answer evaluates to -8.

 

[danago]

I would probably first do polynomial long division to reduce the rational function to a form:

A(x) + B(x)/C(x)

Where A(x), B(x) and C(x) are all polynomials such that the degree of B(x) is less than the degree of C(x), and then just reduce the B(x)/C(x) to its partial fraction parts.

 

[MadmanMurray]

Heres the method I used

Say I have
x+4 / x² - 3x+2

I factorize the bottom line and get
x+4 / (x-1)(x-2)

then I say
x+4 / (x-1)(x-2) = A / (x-1) + B / (x-2)

then I turn that into this
x+4 = A(x-2) + B(x-1)

which is supposed to be true for all values of x so I get my value for A by letting x=1 which will cancel out B.

This is how they thought us how to do it but it doesn't seem to be working;

 

[Delphi51]

Maybe you should look up "partial fractions". It is quite an interesting technique. I found something on it in Wikipedia and was able to find a solution to your problem fairly quickly. There was something about reducing the degree of the denominator by one, so I wrote
x2 - x - 5 ... Ax + B .. Cx + D
--------- = ------- + --------- ... (ignore the dots - trying to align it)
(x-2)(x-1) ... (x-2) ... (x-1)

The expression on the left side is your problem, with a factor of x taken out to make it simpler (you can multiply the answer by x later).
The idea of this technique is to guess that this can be done, then use the above equation to find the values of A, B, C and D that make it work - if you can.

Multiply every term by the common denominator to eliminate fractions. Then play. For example, letting x=0 on both sides gives -5 = -B - 2D.
Another example, on the left the coefficient of x2 is 1 and on the right, A + C, so you have A+C = 1. Do this kind of thing a few times and pretty soon you can figure out what A, B, C and D are. Then you must check to see if your answer equals the original expression!

 

[HallsofIvy]

Heres the method I used

Say I have
x+4 / x² - 3x+2

But your example was not of this form: you had
$$\frac{x^3- x^2- 5x}{x^2- 3x+ 2}$$
which is an "improper fraction"- the numerator is of higher dimension than the denominator. The first thing you should do is divide: $x^2- 3x+ 2$ divides into $x^3- x^2- 5x$ x+ 2 times with remainder $3x- 2$.

As danago said, that reduces the problem to the "mixed fraction"
$$x+ 2+ \frac{3x- 2}{x^2- 3x- 2}$$
Apply partial fractions to that fraction.

I factorize the bottom line and get
x+4 / (x-1)(x-2)

then I say
x+4 / (x-1)(x-2) = A / (x-1) + B / (x-2)

then I turn that into this
x+4 = A(x-2) + B(x-1)

which is supposed to be true for all values of x so I get my value for A by letting x=1 which will cancel out B.

This is how they thought us how to do it but it doesn't seem to be working;

 

[addoow]

Homework Statement

x³ - x² - 5x / x² -3x +2

Homework Equations

The Attempt at a Solution

Heres what I got
5/x-1 + 6/x-2
but if I multiply those 2 fractions together I do not get a cubic equation on top of the line. I learned how to do these questions with linear equations on top I'm lost here though. Is there some trick that I don't know about?

Hello,these type of problems seem daunting at first sight but by using some algebraic manipulations we can easily decompose this polynomial into simpler fractions by using the partial fraction decomposition process.
Lets start

x3 - x2 - 5x / x2 -3x +2=x+2+(-x-4)/(x^2-3x+2) [using algebraic long division]
lets consider on the fractional part that is (-x-4)/(x^2-3x+2)

(-x-4)/(x^2-3x+2) factorise the denominator
(-x-4)/(x-2)(x-1) then next we will decompose them into two parts

(-x-4)/(x-2)(x-1)=A/(x-2)+B/(x-1) (!) which is the same thing as

(-x-4)/(x-2)(x-1)=A/(x-2)+B/(x-1)=A(x-1)+B(x-2)/(x-2)(x-1) then we have(-x-4)/(x-2)(x-1)=A(x-1)+B(x-2)/(x-2)(x-1) we can see that the denominators are equal therefore numerators must also be equal.
Lets set them equal
-x-4=A(x-1)+B(x-2) when x=2 A=-6 and when x=1 B=5 then just substitute these values into the original equation (!)
we have
(-x-4)/(x-2)(x-1)=A(x-1)+B(x-2)/(x-2)(x-1)=-6/(x-2)+5/(x-1) and finally

x3 - x2 - 5x / x2 -3x +2=(x+2)-6/(x-2)+5/(x-1)that is the partial fraction expansion process.