Partial Fraction Decomposion

Summary

This video tutorial explains partial fraction decomposition, a method to break down complex rational functions into simpler fractions. The instructor demonstrates how to decompose fractions with linear and quadratic factors, using examples to clarify the process. Key terms like linear and repeated factors are introduced, and the video shows how to solve for constants in the decomposition through either system of equations or plugging in values. The method is applied to two problems, illustrating the technique's practical use.

Takeaways

• 📝 Partial Fraction Decomposition is a method used to break down a single complex fraction into simpler fractions.
• 📝 The process involves finding a common denominator to combine fractions, which is the reverse of what is done in decomposition.
• 📝 Factoring the denominator completely is crucial, especially focusing on linear and quadratic factors.
• 📝 Linear factors are of the form ax + b, while quadratic factors are of the form ax^2 + bx + c.
• 📝 Repeated factors, whether linear or quadratic, require special notation and affect the decomposition process.
• 📝 The video demonstrates how to decompose a rational function by setting up constants (a, b, etc.) over each factor and solving for these constants.
• 📝 Multiplying both sides of the equation by the denominator is a key step to isolate and solve for the constants.
• 📝 Plugging in specific values for x can simplify the process of finding constants in the decomposition.
• 📝 Once constants are determined, they are substituted back into the original equation to express the decomposed fractions.
• 📝 The final step is to verify the decomposition by combining the fractions to ensure they match the original function.

Q & A

What is partial fraction decomposition?

Partial fraction decomposition is a method used in calculus to break down a complex rational function into simpler fractions. It involves expressing a single fraction as the sum of two or more fractions, each with a simpler denominator.

How does partial fraction decomposition help in solving problems?

It simplifies the process of integrating or differentiating complex rational functions by breaking them down into simpler components that are easier to work with.

What is the first step in partial fraction decomposition?

The first step is to factor the denominator completely, which may involve finding linear and/or quadratic factors.

What is a linear factor in the context of partial fraction decomposition?

A linear factor is a first-degree polynomial, such as x + 2, 3x - 5, or 4x + 8.

Can you give an example of a quadratic factor?

A quadratic factor is a second-degree polynomial that may or may not be factorable, such as x^2 + 8x + 3 or x^2 + 7.

What is a repeated linear factor?

A repeated linear factor is a factor that appears more than once in the factorization of the denominator, such as x^2 (x squared).

How do you determine the constants in partial fraction decomposition?

You can determine the constants by setting up a system of linear equations or by plugging in specific x values to solve for each constant.

Why is it necessary to multiply both sides of the equation by the denominator when performing partial fraction decomposition?

Multiplying both sides by the denominator allows you to clear the fractions and solve for the unknown constants in a simpler manner.

What is the purpose of plugging in specific x values in partial fraction decomposition?

Plugging in specific x values simplifies the equation by eliminating terms and makes it easier to solve for the unknown constants.

How do you verify that your partial fraction decomposition is correct?

You can verify it by combining the decomposed fractions back into a single fraction and checking if it matches the original rational function.

Can you provide a simple example of partial fraction decomposition from the script?

Yes, the script provides an example of decomposing 7x - 23 divided by (x^2 - 7x + 10) into 3/(x - 2) + 4/(x - 5).

Outlines

00:00

📚 Introduction to Partial Fraction Decomposition

This paragraph introduces the concept of partial fraction decomposition, a mathematical technique used to break down a single complex fraction into simpler fractions. The narrator uses an example to illustrate the process, starting with combining two fractions with different denominators into one by finding a common denominator. The example given is (2/(3x+4)) + (5y/15xy), which simplifies to (10y + 12x)/15xy. The narrator then explains that partial fraction decomposition is the reverse process, where a single fraction is decomposed into multiple smaller fractions. The example problem presented is to decompose the fraction (7x - 23)/(x^2 - 7x + 10). The first step is to factor the denominator completely, which in this case is factoring x^2 - 7x + 10 into (x - 2)(x - 5). The narrator then explains the terms linear and quadratic factors and how they affect the decomposition process.

05:00

🔍 Decomposition Process and Solving for Constants

The paragraph continues with the process of partial fraction decomposition by setting up the equation (7x - 23)/[(x - 2)(x - 5)] = a/(x - 2) + b/(x - 5). The goal is to determine the values of constants 'a' and 'b'. The narrator suggests two methods to find these constants: using a system of linear equations or plugging in x values. For simplicity, the narrator opts for the latter method. By substituting x = 5, the equation simplifies to 3b = 12, solving which gives b = 4. Similarly, by substituting x = 2, the equation simplifies to -9 = -3a, solving which gives a = 3. The narrator then verifies the decomposition by combining the fractions and checking if it matches the original fraction. The paragraph concludes with the correct decomposition of the fraction into 3/(x - 2) + 4/(x - 5).

10:02

🔢 Applying the Decomposition to Another Problem

In this paragraph, the narrator applies the partial fraction decomposition technique to another problem: decomposing the fraction (29 - 3x)/(x^2 - x - 6). The first step is to factor the denominator, which is done by finding two numbers that multiply to -6 and add to -1, resulting in the factors (x - 3)(x + 2). The equation is set up as (29 - 3x)/[(x - 3)(x + 2)] = a/(x - 3) + b/(x + 2). The narrator then multiplies both sides by the denominator to isolate 'a' and 'b'. By substituting x = 3, the equation simplifies to 5a = 20, solving which gives a = 4. Similarly, by substituting x = -2, the equation simplifies to -5b = 35, solving which gives b = -7. The final decomposition is presented as 4/(x - 3) - 7/(x + 2), completing the solution to the second problem.

Keywords

💡Partial Fraction Decomposition

Partial Fraction Decomposition is a method in algebra used to break down a complex rational expression into simpler fractions. It is the reverse process of combining fractions with a common denominator. In the video, this technique is used to simplify expressions like '7x - 23' divided by 'x^2 - 7x + 10' into '3/(x-2) + 4/(x-5)'. The process involves factoring the denominator and setting up equations to solve for unknown constants in the numerators of the resulting fractions.

💡Common Denominator

A common denominator is a single denominator that multiple fractions can share, allowing them to be combined into one fraction. In the script, the process of finding a common denominator is illustrated when combining '2/3x + 4/5y' into a single fraction by multiplying the second fraction by '3x/3x' and the first by '5y/5y', resulting in a common denominator of '15xy'.

💡Rational Function

A rational function is a function that is the ratio of two polynomials. The video script discusses decomposing rational functions into partial fractions, specifically the function '7x - 23' divided by 'x^2 - 7x + 10'. Rational functions are integral to calculus and algebra, often appearing in problems involving limits and integrals.

💡Factoring

Factoring is the process of breaking down a polynomial into a product of its factors. In the video, factoring is used to decompose the quadratic trinomial 'x^2 - 7x + 10' into '(x-2)(x-5)'. This step is crucial for partial fraction decomposition, as it allows for the identification of linear factors that will be used to create the simpler fractions.

💡Linear Factors

Linear factors are algebraic expressions of the first degree, such as 'x + 2' or '3x - 5'. The script mentions that when a denominator has linear factors, the corresponding numerator in the partial fraction decomposition should be a constant. For example, '(x-2)' and '(x-5)' are linear factors in the decomposition of the rational function.

💡Quadratic Factors

Quadratic factors are polynomial expressions of the second degree, like 'x^2 + 8x + 3'. In the video, quadratic factors are distinguished from linear factors and are used to determine the form of the numerator in the partial fraction decomposition when the denominator contains quadratic terms.

💡Repeated Linear Factor

A repeated linear factor is a linear factor that appears more than once in a polynomial's factorization. The script uses 'x^2' as an example of a repeated linear factor. When a rational function has a repeated linear factor in the denominator, the numerator in the partial fraction will be a linear expression rather than a constant.

💡Constants

Constants in the context of the video are the unknown values in the numerators of the partial fractions that need to be determined. For instance, in the decomposition of '7x - 23' divided by '(x-2)(x-5)', the constants 'a' and 'b' are solved to be '3' and '4', respectively.

💡System of Linear Equations

A system of linear equations is a set of two or more equations involving the same variables. The video mentions that one method to find the constants in partial fractions is by setting up and solving such a system. However, the script also suggests plugging in x values as a simpler alternative for this particular problem.

💡Plugging in x Values

Plugging in x values is a method to solve for constants in partial fraction decomposition by substituting specific values of x that simplify the equation. The video uses this method to find 'a' and 'b' by choosing x values that make certain terms disappear, such as plugging in 'x = 5' and 'x = 2' to solve for 'a' and 'b'.

💡Verification

Verification in the context of the video is the process of checking the correctness of the partial fraction decomposition by combining the decomposed fractions back into the original rational function. The script demonstrates this by recombining '3/(x-2) + 4/(x-5)' to ensure it equals the original function '7x - 23' divided by '(x-2)(x-5)'.

Highlights

Introduction to partial fraction decomposition

Combining fractions with common denominators

Breaking down a single fraction into smaller fractions

Example problem: 7x - 23 divided by x^2 - 7x + 10

Factoring the denominator completely

Identifying linear and quadratic factors

Understanding repeated linear and quadratic factors

Setting up partial fractions with constants a and b

Multiplying both sides by the denominator to find a and b

Solving for constants a and b by plugging in x values

Calculating a when x equals 2

Calculating b when x equals 5

Final answer for the first example problem

Checking the answer by combining fractions

Second example problem: 29 - 3x divided by x^2 - x - 6

Factoring the denominator of the second problem

Setting up partial fractions for the second problem

Finding a by plugging in x equals 3

Finding b by plugging in x equals -2

Final answer for the second example problem

Transcripts

in this video we're going to go over

partial fraction decomposition i'm going

to show you how to solve many of these

problems but before we do that

let's talk about

what partial fraction decomposition is

so what exactly is it well to illustrate

it let me give you an example problem

let's say if we have 2 over 3x

plus 4

over 5y

how can we combine these two fractions

into one single fraction

well we need to get common denominators

first so i'm going to multiply the

second fraction by 3x over 3x

and the first one

by 5y over 5y

so this will give me 10y

over 15xy

plus

12x

over 15xy

now because

i have a common denominator i can

combine it into a single fraction so i

can write it as 10y

plus 12x

divided by 15xy

now what partial fraction decomposition

allows me to do

is it allows me to take a single

fraction like this one

and break it down into two smaller

fractions

so it's the reverse of combining two

fractions into a single fraction so you

take a single fraction and break it down

into multiple

smaller fractions it could be two

fractions three four it depends on the

nature

of this particular fraction

and so that's the basic idea behind

partial fraction decomposition

so let's work out an example

let's say we have a rational function

7x minus 23

divided by

x squared minus 7x

plus 10.

so how can we take this fraction

and break it down into smaller fractions

the first thing we need to do is we need

to factor this fraction completely

we can't factor the numerator however we

can factor in this trinomial x squared

minus seven x plus ten

so what two numbers multiply to ten

but add to the middle coefficient

negative seven

so we know five times two is ten but

negative five plus negative two adds up

to negative 7.

so we have 7x minus 3

divided by

x minus 2

and x minus 5.

so

you need to write the denominator in

terms of linear factors and quadratic

factors

so what exactly is a linear factor

linear factors are like x plus two

three x minus five

four x plus eight

x

these are linear factors

a quadratic factor

would look something like this

x squared plus

8x

plus 3 which is not factorable

or it could be x squared plus seven

these are

quadratic factors

now there are some other terms that you

need to be familiar with

how would you describe

this term

this is known as a repeated linear

factor

so x squared is also a repeated linear

factor

7x plus 3 squared that's a repeated

linear factor

now x squared plus 1 squared that's a

repeated quadratic factor

be familiar with these terms because

it's going to affect the way

this fraction is decomposed

so we have two linear factors

and what we need to write is two

fractions on this side

and each fraction

will contain a linear factor

now anytime you have a linear factor on

the bottom you need to put a constant on

top

now we're going to choose two different

constants a and b

our goal

is to determine a and b

is a equal to three is be equal to four

we need to figure that out

so what i'm gonna do

is i'm gonna multiply both sides of the

equation

by this denominator that is by

x minus two

and x minus 5.

so if we multiply this fraction by what

we have here

x minus 2 and x minus 5 will cancel

giving us 7x minus 23 on the left side

now if i take this fraction and multiply

by what i have here

the x minus two terms will cancel

leaving behind a

times x minus five

now these two will cancel and i'm gonna

be left with b times x minus two

now when you get to this part

there's two ways in which you could find

a and b

you could use a system of linear

equations

or you can plug in x values and

determine a and b

for a simple problem like this it's best

to plug in x values

so let's try plugging in

x equals five because if we do so

this will disappear five minus five is

zero and zero times a is zero

so then this becomes

seven x minus twenty three or seven

times five

minus twenty three because we need to

replace that with five

that's equal to a

times five minus five

plus b

times five minus two

seven times five is thirty five

and five minus five is zero so this

disappears

and five minus two is three

now thirty five minus twenty three is

twelve

and so that's equal to 3b

and if we divide both sides by 3

b is equal to 4. so i'm just going to

rewrite that here

so now we need to calculate a

so this time

let's plug in 2

so if x is equal to 2 let's see what's

going to happen so we're going to have 7

times 2 minus twenty three

and that's equal to a

times

two minus five

plus

b times two minus two

seven times two is fourteen

and two minus five is negative three

and two minus two is zero

now fourteen minus twenty three

that's a negative nine and that's equal

to negative three a

b times zero is zero so that disappears

and now let's divide both sides by

negative three

so negative nine divided by negative

three is three

and so

that's the value for a

now let's go back to our original

problem

so we had 7x minus 23

divided by

x minus two

times

x minus five

and that was equal to a

over x minus two

plus b

over x minus five

now we have the value of a a is three

and b

is four

so therefore

we could see that this fraction

is equal to three over x minus two plus

four

over x minus five that's the answer

now if you ever need to check it

simply combine the two fractions

and see if they give you

what you started with here so let's go

ahead and try that i'm gonna multiply

this fraction by x minus two

over x minus two

whatever you do to the top you have to

do to the bottom and this one i'm gonna

multiply by x minus five

divided by x minus five

three times x minus five is three x

minus fifteen

and on the bottom we have x minus five

times x minus two

which i'm gonna leave it that way

and then four times x minus two

that's going to be 4x

minus 8

divided by x minus 5x minus 2.

now

if we add 3x and 4x

that will give us

7x and then if we add negative 15 and

negative 8

that will give us a negative 23.

so as you can see

the answer that we had at the beginning

was correct

so it's 3 over x minus 2

plus 4 over x minus 5. that's the answer

now for the sake of practice let's try

another similar problem

so let's say

let me rewrite that if we want to

decompose 29

minus 3x

divided by

x squared

minus x minus 6.

feel free to pause the video go ahead

and try that

so first we need to factor the

denominator by the way notice that the

denominator is always one degree higher

than the numerator

now what two numbers multiplies to

negative six and add to the middle

coefficient negative one

this is going to be negative three and

positive two

negative three plus two adds up to

negative one

so we have 29 minus three x

divided by

x minus 3

times x plus 2.

so let's set this equal to a

over x minus 3

plus b

over x plus 2

since we have two

linear factors

now just like before we're gonna

multiply both sides of this equation

by this denominator

so that is by x minus three times x plus

two

so these two will cancel and on the left

side we're going to get twenty nine

minus three x

now if we take a over x minus 3 and

multiply it by this

we can see that

x minus 3

will cancel

and then we're going to get a times x

plus two

and then if we take this fraction

multiply by x minus three times x plus

two

the x plus two factors will cancel

leaving behind

b

times x minus 3.

actually let's keep that there for now

so

at this point let's plug in x equals

three

this is going to be 29 minus three times

three

and that's equal to a

times three plus two

plus b

times three minus three

three minus three is nine

and three plus two is five

three minus three is zero so this

disappears

29 minus nine is twenty

and that's equal to five a

so if we divide both sides by 5

we can see that a

is equal to 4.

i'm going to put that in the side over

here

now let's plug in negative 2 so that

this term becomes 0.

so we're going to have 29 minus 3 times

negative 2

and that's equal to a

negative 2 plus 2

plus b

times negative two minus three

now

negative three times negative two is

positive six

negative two plus two is zero

and negative two minus 3 is negative 5.

29 plus 6 is 35

and so that's equal to negative 5b

so b is going to be 35

divided by negative 5

which is negative 7.

so now that we have a and b

we can now write the final answer

so let's plug it in to this expression

so 29 over i mean 29 minus 3x over x

squared minus x minus 6

that's equal to 4

divided by x minus 3

and then b is negative 7. so instead of

writing plus negative 7 i could simply

write

minus 7

because a positive number times a

negative number is still a negative

result

so it's going to be minus 7 over x plus

2.

and this is the answer

you