Partial Fraction Decomposition

Partial Fraction Decomposition

Let's first review adding rational expressions...

Notice the common denominator is a product of 2 binomials

Partial fraction decomposition...

Is essentially the reverse!

PFD is the process of taking one fraction and breaking it down into the sum of 2 or more fractions with prime denominators.

Before solving, first check to see:

• If the degree of the numerator is higher than the degree of the denominator, you must long divide first. The remainder is what we will perform PFD on.

2nd: Factor then set up fraction sums:

• Factor the denominator completely so each factor is prime. Choose one of 3 models depending on the type of factors you got in the denominator

• 3 types of factors: Linear, Repeated, Non-factorable

• Create a sum of fractions for each type of factor putting a different letter (representing a constant) over each that we will later solve for

3 Decomposition Models:

• 1) Linear Factors (ax+b form): For each linear factor, create a fraction sum and put a different letter over each.   Ex: if the denom. includes (x+2)(x-3) the model is:  $\frac{A}{x+2}+\frac{B}{x-3}$  

• 2) Repeated Linear Factors: If 2 or more factors are the same, like (px+q)^2, create a fraction for each power of the factor. Ex: if the denom. is (x+3)^2, the model is:  $\frac{A}{\left(x+3\right)^1}+\frac{B}{\left(x+3\right)^2}$  

• 3) Nonfactorable Quadratic: For each non-factorable quadratic (like px^2+q), create a fraction with a linear form numerator. Ex if the denom. is (x^2+x+7), the model is: $\frac{Ax+B}{x^2+x+7}$  

Examples of the model setup:

• Note: models can be mixed depending on the types of factors.  Use a different letter over each.

•  $\frac{x^2+8x+6}{\left(x+5\right)\left(x-4\right)}=\frac{A}{x+5}+\frac{B}{x-4}$  

•  $\frac{x^2+8x+6}{\left(x+5\right)^2\left(x-4\right)}=\frac{A}{\left(x+5\right)}+\frac{B}{\left(x+5\right)^2}+\frac{C}{\left(x-4\right)}$  

•  $\frac{x^2+8x+6}{\left(x^2+5\right)\left(x-4\right)}=\frac{Ax+B}{x^2+5}+\frac{C}{x-4}$  

Multiple Choice

Set up the partial fractions...

Multiple Choice

Which one should be used in the partial fraction decomposition of  $\frac{x-11}{\left(x+3\right)^2\left(x-4\right)}?$  

•  $\frac{A}{\left(x+3\right)}+\frac{B}{\left(x+3\right)}+\frac{C}{\left(x-4\right)}$  

•  $\frac{A}{\left(x-4\right)}+\frac{Bx+C}{\left(x+3\right)^2}$  

•  $\frac{A}{\left(x+3\right)^2}+\frac{B}{\left(x-4\right)}$  

•  $\frac{A}{\left(x+3\right)}+\frac{B}{\left(x+3\right)^2}+\frac{C}{\left(x-4\right)}$  

Multiple Choice

Which one should be used in the partial fraction decomposition of  $\frac{4x}{\left(x+1\right)\left(3+x^2\right)}?$  

•  $\frac{A}{\left(x+1\right)}+\frac{B}{\left(3+x\right)}+\frac{C}{\left(3+x\right)^2}$  

•  $\frac{A}{\left(x+1\right)}+\frac{Bx+C}{\left(3+x^2\right)}$  

•  $\frac{A}{\left(x+1\right)}+\frac{B}{\left(3+x^2\right)}$  

•  $\frac{A}{\left(3+x^2\right)}+\frac{Bx}{\left(1+x\right)}$  

Find the PFD for:  $\frac{7x-8}{\left(2x-1\right)\left(x-2\right)}$  

The denominator has already been factored.  So how should the model fraction sum look?

Multiple Choice

Which is the correct setup for:  $\frac{7x-8}{\left(2x-1\right)\left(x-2\right)}=?$  

•  $\frac{Ax}{2x-1}+\frac{Bx}{x-2}$  

•  $\frac{A}{2x-1}-\frac{B}{x-2}$  

•  $\frac{A}{2x-1}+\frac{B}{x-2}$  

 $\frac{7x-8}{\left(2x-1\right)\left(x-2\right)}=\frac{A}{2x-1}+\frac{B}{x-2}$  

Did you get this?  Now we will solve for the constants of A and B.

First, we multiply each term by the LCD.  Just like solving a rational equation.

 Note: the LCD will always be the denominator of the original fraction.

Multiple Choice

Multiply each term by the LCD of (2x-1)(x-2)  Which is the result?

 $\frac{7x-8}{\left(2x-1\right)\left(x-2\right)}=\frac{A}{2x-1}+\frac{B}{x-2}$ 

•  $7x-8=A\left(x-2\right)^2+B\left(2x-1\right)^2$  
  

•  $7x-8=A\left(x-2\right)+B\left(2x-1\right)$  

•  $7x-8=A\left(2x-1\right)+B\left(x-2\right)$  

You should have gotten: $7x-8=A\left(x-2\right)+B\left(2x-1\right)$  

• Notice we have too many variables to try to find A or B.

• Let's try to eliminate the A or B by plugging in a value of x that will do that.

• To 'eliminate' A we could plug in x = 2 because A(2-2) = 0.  Then we can solve for B

Multiple Choice

Let x = 2.  What is B?

   $7x-8=A\left(x-2\right)+B\left(2x-1\right)$  

• B = 3

• B = 2

• B = 0

• B = 1

Multiple Choice

To 'eliminate' B (so we can solve for A), let x = 1/2.  What is the A value?

   $7x-8=A\left(x-2\right)+B\left(2x-1\right)$  

• A = 3

• A = 2

• A = 0

• A = 1

The final answer 

Replace the A and B in our model to get the final answer of:  $\frac{3}{2x-1}+\frac{2}{x-2}$  

Here's a recap

• Write out the sum of fractions based on the factors of the denominator

• Multiply each term by the LCD

• Plug in values of x that 'eliminate' each of the constants A, B, etc.

• Replace into the fraction model

Repeated Factors

• Solve the same way we did Linear Factors:

• Multiply by the LCD

• Plug in a value of x that eliminates one of the constants, A, B, etc.

• When there are no other 'convenient' values of x to use, plug in the constant value (A, B, etc.) you already found and pick any other x

Multiple Choice

Which is the correct set-up for the PFD?    $\frac{x-4}{\left(x-2\right)^2}$ 

•  $=\frac{A}{x-2}+\frac{B}{\left(x-2\right)^2}$  

•  $=\frac{A}{x-2}+\frac{B}{\left(x-2\right)}$  

•  $=\frac{Ax+B}{x-2}+\frac{C}{\left(x-2\right)^2}$  

Multiple Choice

You should have gotten:   $\frac{x-4}{\left(x-2\right)^2}=\frac{A}{x-2}+\frac{B}{\left(x-2\right)^2}$ .

Now multiply by the LCD.  Which is the correct expression?

•  $x-4=A+B\left(x-2\right)$  

•  $x-4=A\left(x-2\right)^2+B\left(x-2\right)$  

•  $x-4=A\left(x-2\right)+B$  

Multiple Choice

You should have gotten:      $x-4=A\left(x-2\right)+B$ .

Finish solving for A and B.

•  $=\frac{-1}{x-2}+\frac{2}{\left(x-2\right)^2}$  

•  $=\frac{1}{x-2}+\frac{-2}{\left(x-2\right)^2}$  

•  $=\frac{-2}{x-2}+\frac{1}{\left(x-2\right)^2}$  

Here's a recap...

Non-factorable Quadratic

• Instead of 'eliminating' a constant value, we multiply everything out

• Set up little equations that equate the left terms (x^2, x's, and constants) with their equivalent terms on the right and solve.

• ***in this example you may want to plug in x = 2 in the 2nd step first so you can solve for C to use later.

Example:

• Notice how the model includes both a linear and a non-factorable quadratic

• Next, multiply through by the LCD

• Then distribute and set up equations

Multiple Choice

What is the result when you multiply by the LCD?

 $\frac{x+4}{x\left(x^2-2\right)}=\frac{A}{x}+\frac{Bx+C}{x^2-2}$  

•  $x+4=A\left(x^2-2\right)+\left(Bx+Cx\right)$  

•  $x+4=A\left(x\right)+\left(Bx+C\right)\left(x^2-2\right)$  

•  $x+4=A\left(x^2-2\right)+\left(Bx+C\right)\left(x\right)$  

Multiple Choice

What would be the correct distribution?

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

•  $x+4=Ax^2-2Ax+Bx^2+Cx^2$  

•  $x+4=Ax^2-2A+Bx^2+Cx$  

•  $x+4=Ax^2-2A+Bx^2+Cx^2$  

Multiple Select

Which 3 equations should be set up in order to solve for A, B, and C?
(Hint: match degree terms)

 $x+4=Ax^2-2A+Bx^2+Cx$ 

•  $4\ =\ -2A$  

•  $x\ =\ Cx^2$  

•  $0x^2=Ax^2+Bx^2$  

•  $x=Cx$  

•  $x=-2A$  

Multiple Choice

Now use these to solve for A, B, and C.  Ignore the x^2 and x terms and just equate the coefficients:

 $0x^2=Ax^2+Bx^2$  (OR  $0=A+B$ )  
 $x=Cx$  (OR   $1=C$ ) 
 $4=-2A$  

• A = 0, B = 2, C = 1

• A = -2, B =0, C = 2

• A = -2, B = 2, C = 1

Final answer:

 $\frac{x+4}{x\left(x^2-2\right)}\equiv\frac{-2}{x}+\frac{2x+1}{x^2-2}$  

That's all folks!  Now go practice...