11.5 Partial Frac. Decomp Case 1 & 2

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Mathematics

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9th - 12th Grade

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Teacher karp

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16 Slides • 10 Questions

11.5 Partial Fraction Decomposition

Case 1; non-repeat linear factors
&

Case 2; Repeating linear factors

These factors are in the denominator

We first start with proper rational expression. Notice : Degree in the numertor is less than the degree in the denominator.

Multiple Choice

Which of the following is an proper rational equation?

(x^2 + 1)/(2x)

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

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

(x - 1)/(x + 2)

Multiple Select

Okay here is a bit of a different direction.

Select all that apply......

If you had the fraction 5/7 which of these could be used to split up the fraction?

(2/7)+(3/7)

(1/7)+(4/7)

(8/7)-(3/7)

What if we wanted to do the same to this proper rational function?

First Factor the denominator as you see below.

Now set up the equation but we do not know what will be in the numerator so we will use a different variable.

You Try! Make sure you have paper out so you can write down steps BEFORE you go to the next slide. Factor First and go to next slide

Check out the right side of the equation

This a good start.

I know you got this...

Now write this as an equation...by doing the following....

Place a capital letter A over one of the nonrepeating factors of x and B over the other non-repeating factor.

you should have...

you got this right?

What is left?

Here is the next NEW step.
Fraction "Bust" the entire equation by multiplying the ENTIRE equation by x(x-1)

distribute A to get

Regroup the x terms with x terms
and the constant terms with constant terms...

You now have a system of equations. Well almost...

If you take the first line and divide by x from both sides of the =, you can see that we get the system below

If we back sub A=2 into the first line we get B=-1.
So what will our fractions look like now?

Multiple Choice

What should your two fractions look like now? if A= 2 and B= -1;

Recall we started with the equation below

$\frac{x-2}{x^2-x}=\frac{A}{x}+\frac{B}{x-1}$

$\frac{2}{x}+\frac{-1}{x-1}$

$\frac{2}{x}+\frac{1}{x-1}$

$\frac{-2}{x}+\frac{1}{x-1}$

$\frac{2}{x-1}+\frac{-1}{x}$

Multiple Choice

Since the expression below is PROPER; performing partial fraction decomposition what is the initial set up?

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

$\frac{A}{x+1}+\frac{B}{x-2}$

$\frac{A}{x}+\frac{B}{x+1}$

$\frac{A}{x+1}+\frac{B}{x}$

Multiple Choice

Determine the first step for partial fraction decomposition of

$\frac{2x-1}{\left(x+1\right)\left(x-3\right)}$

$\frac{A}{x+1}+\frac{B}{x+3}$

$\frac{A}{x+1}+\frac{B}{x-3}$

$\frac{A}{x-1}+\frac{B}{x+3}$

$\frac{A}{x-1}+\frac{B}{x-3}$

Multiple Choice

using partial fraction decomposition what is the initial set up?

$\frac{x-1}{x^2+6x+8}$

$\frac{A}{x+2}+\frac{B}{x+4}$

$\frac{A}{x+2}+\frac{B}{x+4}+\frac{C}{x-1}$

$\frac{A}{x+2}+\frac{B}{x-1}$

$\frac{A}{x+2}+\frac{Bx}{x+4}$

Let's look at case 2

Linear repeating factors (in the denominator)

Multiple Choice

using partial fraction decomposition what is the initial set up for this?

$\frac{x^2+x+5}{\left(x+1\right)^2\left(x-1\right)}$

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

$\frac{A}{\left(x+1\right)}+\frac{B}{\left(x+1\right)}+\frac{C}{\left(x-1\right)}$

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

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

Multiple Choice

using partial fraction decomposition what is the initial set up for this?

\frac{x+2}{x\left(x^2+8x+16\right)}

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

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

Partial Fraction Decomposition

Case 1 means we have linear FACTORS in the denominator that do NOT repeat
Case 2 means we have linear FACTORS in the denominator that DO repeat
you can have both in any particular rational function
Initial set up, numerators are just a constant letter like A, B, C, D etc.

Multiple Choice

Finish this problem by writing the partial fraction decomposition

\frac{x-1}{x^2+6x+8}

$\frac{-\frac{3}{2}}{x+2}+\frac{\frac{5}{2}}{x+4}$

$\frac{-\frac{5}{2}}{x+2}+\frac{\frac{3}{2}}{x+4}$

$\frac{\frac{5}{2}}{x+2}+\frac{-\frac{3}{2}}{x+4}$

Multiple Choice

Finish this problem by writing the partial fraction decomposition

\frac{x+2}{x\left(x^2+8x+16\right)}

$\frac{\frac{1}{8}}{x}+\frac{-\frac{1}{8}}{x+4}+\frac{\frac{1}{2}}{\left(x+4\right)^2}$

$\frac{-\frac{1}{8}}{x}+\frac{+\frac{1}{8}}{x+4}+\frac{\frac{1}{2}}{\left(x+4\right)^2}$

$\frac{\frac{1}{2}}{x}+\frac{-\frac{1}{8}}{x+4}+\frac{\frac{1}{8}}{\left(x+4\right)^2}$

Notice that ! IN THE NUMERATOR, we never had x-value in the initial set up?

Case 3 & 4 we will...coming up!

11.5 Partial Fraction Decomposition

Case 1; non-repeat linear factors
&

Case 2; Repeating linear factors

These factors are in the denominator

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11.5 Partial Frac. Decomp Case 1 & 2

11.5 Partial Fraction Decomposition

Case 1; non-repeat linear factors &

Case 2; Repeating linear factorsThese factors are in the denominator

What if we wanted to do the same to this proper rational function?

First Factor the denominator as you see below.

Now set up the equation but we do not know what will be in the numerator so we will use a different variable.

You Try! Make sure you have paper out so you can write down steps BEFORE you go to the next slide. Factor First and go to next slide

Check out the right side of the equation

This a good start.

I know you got this...

Now write this as an equation...by doing the following....

Place a capital letter A over one of the nonrepeating factors of x and B over the other non-repeating factor.

If you take the first line and divide by x from both sides of the =, you can see that we get the system below

If we back sub A=2 into the first line we get B=-1. So what will our fractions look like now?

Let's look at case 2

Partial Fraction Decomposition

Notice that ! IN THE NUMERATOR, we never had x-value in the initial set up?

11.5 Partial Fraction Decomposition

Case 1; non-repeat linear factors &

Case 2; Repeating linear factorsThese factors are in the denominator

Similar Resources on Wayground

Popular Resources on Wayground

Discover more resources for Mathematics

Case 1; non-repeat linear factors
&

Case 2; Repeating linear factors

These factors are in the denominator

If we back sub A=2 into the first line we get B=-1.
So what will our fractions look like now?

Case 1; non-repeat linear factors
&

Case 2; Repeating linear factors

These factors are in the denominator