4.6 Rational Equations and Partial Fraction Decomposition.

4.6

Rational Equations and Partial Fraction Decomposition

A rational expression is a fraction with polynomials for the numerator and denominator.

1

3 and ,

3

2 ,

1 2

x

x

xxare rational expressions. For example,

If x is replaced by a number making the denominator of a rational expression zero, the value of the rational expression is undefined.

Example: Evaluate for x = –3, 0, and 1.1

9 and ,

3

2 ,

1 2

x

x

xx

xx

1

3

2

x 1

92

x

x

undefined1 12

1

33

1 0undefined

03

2undefined 9

A rational equation is an equation between rational expressions.

For example, and are rational equations.3

21

xx xx

x

x

x

2

1

1

33

2

4. Check the solutions. 3. Solve the resulting polynomial equation.

2. Clear denominators by multiplying both sides of the equation by the LCM.

1. Find the LCM of the denominators.

To solve a rational equation:

Examples: 1. Solve: .3

1

3

1

x

x

xFind the LCM.

Multiply by LCM = (x – 3).

Solve for x.

LCM = x – 3.

1 = x + 1x = 0

Check. Substitute 0.

Simplify.

3

1

3

1

(0)

(0)(0)

3

1

3

1

True.

2. Solve: .1

21

xx

x – 1 = 2x

Find the LCM.LCM = x(x – 1).

Multiply by LCM.

Simplify.

x = –1 Solve.

1

2

1

xx)1( )1( xxxx

After clearing denominators, a solution of the polynomial equation may make a denominator of the rational equation zero.

Since x2 – 1 = (x – 1)(x + 1),

Since – 1 makes both denominators zero, the rational equation has no solutions.

Example: Solve: .1

1

1

132

xx

x

2x = – 2 x = – 1

3x + 1 = x – 1

Check.

It is critical to check all solutions.

In this case, the value is not a solution of the rational equation.

LCM = (x – 1)(x + 1).

1

1

1

132 xx

x)1)(1( )1)(1( xxxx

Example: Solve: .158

6

3 2

xxx

x

Factor.

Polynomial Equation.

Simplify.

Factor.

The LCM is (x – 3)(x – 5).x2 – 8x + 15 = (x – 3)(x – 5)

x(x – 5) = – 6

x2 – 5x + 6 = 0

(x – 2)(x – 3) = 0

x = 2 or x = 3Check. x = 2 is a solution.Check. x = 3 is not a solution since both sides would be undefined.

158

6

3 2

xxx

xOriginal Equation.

158

6

3 2

xxx

x)5)(3( )5)(3( xxxx

• Sometimes we need more tools to help with rational expressions…

• We will learn to perform a process known as partial fraction decomposition…

To find partial fractions for an expression, we need to reverse the process of adding fractions.

We will also develop a method for reducing a fraction to 3 partial fractions.

)1)(2(

4233

xx

xx

We’ll start by adding 2 fractions.

1

2

2

3

xxe.g.

)1)(2( xx

)1(3 x )2(2 x

)1)(2(

15

xx

x

The partial fractions for are)1)(2(

15

xx

x

1

2

2

3

xx

The expressions are equal for all values of x so we have an identity.

The identity will be important for finding the values of A and B.

)1)(2(

15

xx

x12

x

B

x

A

To find the partial fractions, we start with

2x

A )1)(2( xx

1x

B )1)(2( xx

)1)(2(

15

xx

x )1)(2( xx

)1)(2(

15

xx

x12

x

B

x

A

Multiply by the denominator of the l.h.s.

)1(xA 15xSo,

If we understand the cancelling, we can in future go straight to this line from the 1st line.

To find the partial fractions, we start with

)2( xB

This is where the identity is important.

)2()1(15 xBxAx

The expressions are equal for all values of x, so I can choose to let x = 2.

Why should I choose x = 2 ?

ANS: x = 2 means the coefficient of B is zero, so B disappears and we can solve for A.

A 3

This is where the identity is important.

)2()1(15 xBxAx

2x

What value would you substitute next ?

ANS: Any value would do but x = 1 is good.

)22()12(1)2(5 BAA39

The expressions are equal for all values of x, so I can choose to let x = 2.

This is where the identity is important.

)2()1(15 xBxAx

2x )22()12(1)2(5 BAA39

1x )21()11(1)1(5 BAB36

)1)(2(

15

xx

xSo,

The expressions are equal for all values of x, so I can choose to let x = 2.

If we chose x = 1 instead, we get 4 = 2A – B, giving the same result.

A 3

B 2

12

xxA B

This is where the identity is important.

)2()1(15 xBxAx

2x )22()12(1)2(5 BAA39

1x )21()11(1)1(5 BAB36

)1)(2(

15

xx

xSo,

The expressions are equal for all values of x, so I can choose to let x = 2.

If we chose x = 1 instead, we get 4 = 2A – B, giving the same result.

12

xxAA B

A 3

B 2

This is where the identity is important.

)2()1(15 xBxAx

2x )22()12(1)2(5 BAA39

1x )21()11(1)1(5 BAB36

)1)(2(

15

xx

xSo,

The expressions are equal for all values of x, so I can choose to let x = 2.

If we chose x = 1 instead, we get 4 = 2A – B, giving the same result.

312

xx

2

A 3

B 2

)1)(3(

1

xxSolution: Let

13

x

B

x

A

Multiply by : )1)(3( xx

1 )1(xA

3x )2(1 A

1x )2(1 B21 A

21 B

)3( xBIt’s very important to write this each

time

e.g. 2 Express the following as 2 partial fractions.

)1)(3(

1

xx

We never leave fractions piled up like this, so

• The “halves” are written in the denominators ( as 2s ) and

• the minus sign is moved to the front of the 2nd fraction.

13)1)(3(

1 21

21

xxxxSo,

)1)(3(

1

xx )1(2

1

)3(2

1

xxFinally, we need to check the

answer.A thorough check would be to reverse the process and put the fractions together over a common denominator.

)1)(3(

1

xx )1()3(

x

B

x

AAnother check is to use the “cover-up” method:

We get

)13)(3(

1

xTo check B, substitute x = in the l.h.s. but cover-up)1( x

2

1

Cover-up on the l.h.s. and substitute x = 3 into the l.h.s. only

)3( x

2

1

)( A

)( B )1)(31(

1

x

To check A, find the value of x that makes the factor under A equal to zero( x = 3 )

The method we’ve used finds partial fractions for expressions I’ll call Type 1

where, the denominator has 2 linear factors,

2)12)(3( x x

e.g.x

where, • the denominator has 2 linear factors,

The method we’ve used finds partial fractions for expressions I’ll call Type 1

2)12)(3( x x

e.g.x

( we may have to factorise to find them )

• and the numerator is a polynomial of lower degree than the denominator

The degree of a polynomial is given by the highest power of x.

The method we’ve used finds partial fractions for expressions I’ll call Type 1

2)12)(3( x x

e.g.x

where, • the denominator has 2 linear factors,

The degree of a polynomial is given by the highest power of x.

Here the numerator is of degree

The method we’ve used finds partial fractions for expressions I’ll call Type 1

2)12)(3( x x

e.g.x

1and the denominator of degree

• and the numerator is a polynomial of lower degree than the denominator

where, • the denominator has 2 linear factors,

where, the denominator has 2 linear factors,

The degree of a polynomial is given by the highest power of x.

and the numerator is a polynomial of lower degree then the denominator

The method we’ve used finds partial fractions for expressions I’ll call Type 1

2)12)(3( x x

e.g.x

and the denominator of degree

2Here the numerator is of degree

1

where, the denominator has 2 linear factors,

The degree of a polynomial is given by the highest power of x.

and the numerator is a polynomial of lower degree then the denominator

The method we’ve used finds partial fractions for expressions I’ll call Type 1

2)12)(3( x x

e.g.x

and the denominator of degree

2Here the numerator is of degree

1

SUMMARYTo find partial fractions for expressions

like)12)(3(

2

xx

x

Let

123)12)(3(

2

x

B

x

A

xx

x

• Multiply by the denominator of the l.h.s.• Substitute a value of x that makes the coefficient of B equal to zero and solve for A.• Substitute a value of x that makes the coefficient of A equal to zero and solve for B.• Check the result by reversing the method or using the “cover-up” method.

Express each of the following in partial fractions.

1.

Exercises

)2)(3(

55

xx

x 2.)3)(12(

5

xx

3.

1

22 x

4.)1(

37

xx

x