another fraction decomposition

**cityismine** · May 27th 2008, 07:36 PM

x^2 / (x^3-8)

**Reckoner** · May 27th 2008, 07:55 PM

Originally Posted by cityismine

x^2 / (x^3-8)

Can you factor the denominator? Hint: it's a difference of cubes.

**cityismine** · May 27th 2008, 08:14 PM

Still can't solve it, stuck!

**Jen** · May 27th 2008, 08:25 PM

Originally Posted by cityismine

x^2 / (x^3-8)

Originally Posted by Reckoner

Can you factor the denominator? Hint: it's a difference of cubes.

As Reckoner said, use the difference of cubes

The difference of cubes is,

$a^3-b^3=(a-b)(a^2+ab+b^2)$

and $8=2^3$

Hope that helps.

**cityismine** · May 27th 2008, 08:35 PM

I understand the difference of cubes, but I'm having trouble decomposing the fraction.

**Reckoner** · May 27th 2008, 09:05 PM

Originally Posted by cityismine

I understand the difference of cubes, but I'm having trouble decomposing the fraction.

Show us what you have done so far so that we can see what, exactly, you are having trouble with.

Once you have factored the denominator completely, set up your new fractions:

For every linear factor $(ax + b)$ include the term $\frac{A}{ax + b}$ . If you have powers of the same factor, $(ax + b)^n$ , include all of the terms $\frac{A_1}{ax + b} + \frac{A_2}{(ax + b)^2} + \cdots + \frac{A_n}{(ax + b)^n}$ .

For each quadratic factor (or power of a quadratic factor), $(ax^2 + bx + c)^n$ , include all of the terms $\frac{A_1x + B_1}{ax^2 + bx + c} + \frac{A_2x + B_2}{(ax^2 + bx + c)^2} + \cdots + \frac{A_nx + B_n}{(ax^2 + bx + c)^n}$ .

So, in this case we have

$\frac{x^2}{x^3 - 8} = \frac{x^2}{(x - 2)(x^2 + 2x + 4)}$

$=\frac{x^2}{{\color{red}(x - 2)}{\color{blue}(x^2 + 2x + 4)}}$

$={\color{red}\frac A{x - 2}} + {\color{blue}\frac{Bx + C}{x^2 + 2x + 4}}$

So, we have

$\frac{x^2}{x^3 - 8}=\frac A{x - 2} + \frac{Bx + C}{x^2 + 2x + 4}$

Multiply through by $x^3 - 8$ :

$x^2 = A(x^2 + 2x + 4) + (Bx + C)(x - 2)$

Collect terms of the same degree, and equate the coefficients of both sides. Can you finish?

**Chris L T521** · May 27th 2008, 09:14 PM

Originally Posted by Reckoner

Show us what you have done so far so that we can see what, exactly, you are having trouble with.

Once you have factored the denominator completely, set up your new fractions:

For every linear factor $(ax + b)$ include the term $\frac{A}{ax + b}$ . If you have powers of the same factor, $(ax + b)^n$ , include all of the terms $\frac{A_1}{ax + b} + \frac{A_2}{(ax + b)^2} + \cdots + \frac{A_n}{(ax + b)^n}$ .

For each quadratic factor (or power of a quadratic factor), $(ax^2 + bx + c)^n$ , include all of the terms $\frac{A_1x + B_1}{ax^2 + bx + c} + \frac{A_2x + B_2}{(ax^2 + bx + c)^2} + \cdots + \frac{A_nx + B_n}{(ax^2 + bx + c)^n}$ .

So, in this case we have

$\frac{x^2}{x^3 - 8} = \frac{x^2}{(x - 2)(x^2 + 2x + 4)}$

$=\frac{x^2}{{\color{red}(x - 2)}{\color{blue}(x^2 + 2x + 4)}}$

$={\color{red}\frac A{x - 2}} + \boxed{{\color{blue}\frac B{x^2 + 2x + 4}}}$

So, we have

$\frac{x^2}{x^3 - 8}=\frac A{x - 2} + \boxed{\frac B{x^2 + 2x + 4}}$

Multiply through by $x^3 - 8$ :

$x^2 = A(x^2 + 2x + 4) + B(x - 2)$

Collect terms of the same degree, and equate the coefficients of both sides. Can you finish?

The degree of the numerator is one less than the denominator. It should be $\frac{Bx+C}{x^2+2x+4}$ , not $\frac{B}{x^2+2x+4}$

**Reckoner** · May 27th 2008, 09:18 PM

Originally Posted by Chris L T521

The degree of the numerator is one less than the denominator. It should be $\frac{Bx+C}{x^2+2x+4}$ , not $\frac{B}{x^2+2x+4}$

You're right. I even mentioned as much a few lines above. I should get some rest!

Thread: another fraction decomposition

LinkBack

Thread Tools

Display

another fraction decomposition

Similar Math Help Forum Discussions

Partial Fraction Decomposition

fraction decomposition..??

partial fraction decomposition

fraction decomposition

Fraction Decomposition

Search Tags