1. ## Partial Fractions

Partial Fraction Decomposition

Write the partial fraction decomposition of the rational expression.

x/(x^2-5x+6)

(7x^2-x-12)/x(x+1)(x-1)

(9x^2-x-14)/(x^3-x)

Write the partial fraction decomposition of the rational expression.

(6x+5)/(x-9)^2

(5x^2-5x+7)/(x-1)^3

(x+4)/(x^3-2x^2+x)

Write the partial fraction decomposition of the rational expression.

(12x+3)/(x^3-1)

(12^2-12x+6)/(x-3)(x^2+4)

Write the form of the partial fraction decomposition of the rational expression. It is not necessary to solve for the constants.

(7x-3)/(x^2+x-7)^2

(2x+1)/(x-6)(x^2+x-4)^2

Write the partial fraction decomposition of the rational expression.

(x^2+4x-1)/(x^2+3)^2

(4x^3+4x^2)/(x^2+5)^2

2. Originally Posted by soly_sol
Partial Fraction Decomposition

Write the partial fraction decomposition of the rational expression.

x/(x^2-5x+6)

(7x^2-x-12)/x(x+1)(x-1)

(9x^2-x-14)/(x^3-x)
The trick is:
If you have a factor of $\displaystyle ax + b$ in the denominator you have a term $\displaystyle \frac{A}{ax + b}$ in your decomposition.

If you have a factor of $\displaystyle ax^2 + bx + c$ in the denominator you have a term $\displaystyle \frac{Ax + B}{ax^2 + bx + c}$ in your decomposition.

etc.

If you have factor of $\displaystyle (ax + b)^2$ in your denominator then you have terms $\displaystyle \frac{A}{ax + b} + \frac{B}{(ax + b)^2}$ in your decomposition.

If you have factor of $\displaystyle (ax + b)^3$ in your denominator then you have terms $\displaystyle \frac{A}{ax + b} + \frac{B}{(ax + b)^2} + \frac{C}{(ax + b)^3}$ in your decomposition.

etc.

As an example, let's do the first one:
$\displaystyle \frac{x}{x^2-5x+6} = \frac{x}{(x -2)(x - 3)}$

We have linear factors of x - 2 and x - 3 in the denominator, so we use
$\displaystyle \frac{x}{(x -2)(x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3}$

How do we find A and B? My favorite method (there are many) is to add the fractions on the LHS:
$\displaystyle \frac{x}{(x -2)(x - 3)} = \frac{A(x - 3) + B(x - 2)}{(x - 2)(x - 3)} = \frac{(A + B)x + (-3A - 2B)}{(x - 2)(x - 3)}$
and match the coefficients of powers of x on both sides of the equation.

So we know that
$\displaystyle A + B = 1$
and
$\displaystyle -3A - 2B = 0$

I get that A = -2 and B = 3, thus
$\displaystyle \frac{x}{x^2-5x+6} = -\frac{2}{x - 2} + \frac{3}{x - 3}$

It might take longer than some of the other tricks, but I find it keeps me more organized with what I am doing.

Try the other 2 in this set. I got:
$\displaystyle \frac{7x^2-x-12}{x(x+1)(x-1)} = \frac{12}{x} - \frac{3}{x - 1} - \frac{2}{x + 1}$
and
$\displaystyle \frac{9x^2-x-14}{x^3-x} = \frac{14}{x} - \frac{3}{x - 1} - \frac{2}{x + 1}$

-Dan

3. Originally Posted by soly_sol
x/(x^2-5x+6)
Most of the times I try to avoid partial fractions

Here's a slightly variation:

$\displaystyle x^2-5x+6=(x-2)(x-3).$

We have to manipulate the numerator such that this be equal to $\displaystyle x.$ Note that (tricky)

$\displaystyle 3(x-2)-2(x-3)=3x-6-2x+6=x.$ So, with this we're are done, and

$\displaystyle \frac{x} {{x^2 - 5x + 6}} = \frac{{3(x - 2) - 2(x - 3)}} {{(x - 2)(x - 3)}} = \frac{3} {{x - 3}} - \frac{2} {{x - 2}},$

4. (7x^2-x-12)/x(x+1)(x-1)

(9x^2-x-14)/(x^3-x)

Can you please explain to me how to these? I am having some trouble starting out. For the first one could I make the denomintator x(x-1)^2 since its a difference of squares?

5. Originally Posted by soly_sol
(7x^2-x-12)/x(x+1)(x-1)
set $\displaystyle \frac {7x^2 - x - 12}{x(x + 1)(x - 1)} = \frac Ax + \frac B{x + 1} + \frac C{x - 1}$

(9x^2-x-14)/(x^3-x)
set $\displaystyle \frac {9x^2 - x - 14}{x^3 - x} = \frac {9x^2 - x - 14}{x \left( x^2 - 1 \right)} = \frac {9x^2 - x - 14}{x(x + 1)(x - 1)} = \frac Ax + \frac B{x + 1} + \frac C{x - 1}$

6. Am I doing this right so far:

for the first one: 7x^2-x-12=a(x+1)(x-1)+Bx(x-1)+Cx(x+1)

then for the second one: 9x^2-x-14=a(x+1)(x-1)+Bx(x-1)+Cx(x+1)

7. Originally Posted by soly_sol
Am I doing this right so far:

for the first one: 7x^2-x-12=a(x+1)(x-1)+Bx(x-1)+Cx(x+1)

then for the second one: 9x^2-x-14=a(x+1)(x-1)+Bx(x-1)+Cx(x+1)
yes