12 rational expressions

Rational Expressions

Rational ExpressionsPolynomials are expressions of the form anxn + an-1xn-1+ an-2xn-2+…+ a1x1+ a0 where the a’s are numbers.


For example, 7x12 – 4x5 + 6x2 – 5,

Polynomials are expressions of the form anxn + an-1xn-1+ an-2xn-2+…+ a1x1+ a0 where the a’s are numbers.


For example, 7x12 – 4x5 + 6x2 – 5, πx – 0.762 are polynomials.




Rational (fractional) expressions are expressions of the

form , where P and Q are polynomials.PQ







All polynomials are rational expressions by viewing P as .P1







x – 2 x2 – 2 x + 1 ,







x – 2 x2 – 2 x + 1 ,

x(x – 2) (x + 1) (2x + 1) are rational expressions.







x – 2 x2 – 2 x + 1 ,


x – 2 2 x + 1

is not a rational expression because the

denominator is not a polynomial.







x – 2 x2 – 2 x + 1 ,


x – 2 2 x + 1

is not a rational expression because the

denominator is not a polynomial.

Rational expressions are expressions that describe calculation procedures that involve division (of polynomials).

Rational ExpressionsJust as polynomials may be given in the expanded form or factored form, rational expressions may also be given in these two forms.


The rational expression x2 – 4 x2 + 2x + 1 is in the expanded form.


The rational expression x2 – 4

(x + 2)(x – 2) (x + 1)(x + 1) .

is in the expanded form.

In the factored form, it’s

x2 + 2x + 1


The rational expression x2 – 4 x2 + 2x + 1 (x + 2)(x – 2) (x + 1)(x + 1) .

x – 2 x2 + 1 The expression is in both forms.



Example A. Put the following expressions in the factored form.

a. x2 – 3x – 10 x2 – 3x

b. x2 – 3x + 10 x2 – 3






(x – 5)(x + 2)a. x(x – 3)

x2 – 3x – 10 x2 – 3x

=

b. x2 – 3x + 10 x2 – 3







(x – 5)(x + 2)a. x(x – 3)

x2 – 3x – 10 x2 – 3x

=

b. x2 – 3x + 10 x2 – 3

is in the factored form







(x – 5)(x + 2)a. x(x – 3)

x2 – 3x – 10 x2 – 3x

=

b. x2 – 3x + 10 x2 – 3

is in the factored form

Note that in b. the entire (x2 – 3x + 10) or (x2 – 3) are viewed as a single factors because they can’t be factored further.


We use the factored form to 1. solve equations


We use the factored form to 1. solve equations 2. determine the domain of rational expressions


We use the factored form to 1. solve equations 2. determine the domain of rational expressions2. evaluate given inputs


We use the factored form to 1. solve equations 2. determine the domain of rational expressions2. evaluate given inputs3. determine of the signs of the outputs



Solutions of Equations



Solutions of Equations The solutions of the equation of the form = 0 are the

zeroes of the numerator, so they are the solutions of P = 0.


PQ



Example B. a. Write in the factored form and

solve the equation

The solutions of the equation of the form = 0 are the



x3 + x2 – 2xx2 + 4x + 3

x3 + x2 – 2xx2 + 4x + 3 = 0

PQ




solve the equation




x3 + x2 – 2xx2 + 4x + 3

x3 – 2x2 + 3xx2 + 4x + 3

x3 + x2 – 2xx2 + 4x + 3 = 0

Factor, we get

PQ

=




solve the equation




x3 + x2 – 2xx2 + 4x + 3

x3 – 2x2 + 3xx2 + 4x + 3 = x(x + 2)(x – 1)

(x + 3)(x + 1)

x3 + x2 – 2xx2 + 4x + 3 = 0

Factor, we get

PQ




solve the equation




x3 + x2 – 2xx2 + 4x + 3

x3 – 2x2 + 3xx2 + 4x + 3 = x(x + 2)(x – 1)

(x + 3)(x + 1)

x3 + x2 – 2xx2 + 4x + 3 = 0

Factor, we get

Hence for x3 + x2 – 2xx2 + 4x + 3 = 0, it must be that x(x + 2)(x – 1) = 0

PQ




solve the equation




x3 + x2 – 2xx2 + 4x + 3

x3 – 2x2 + 3xx2 + 4x + 3 = x(x + 2)(x – 1)

(x + 3)(x + 1)

x3 + x2 – 2xx2 + 4x + 3 = 0

Factor, we get

Hence for x3 + x2 – 2xx2 + 4x + 3 = 0, it must be that x(x + 2)(x – 1) = 0

or that x = 0, –2, 1.

PQ

DomainThe domain of a formula is the set of all the numbers that we may use as input values for x.


DomainThe domain of a formula is the set of all the numbers that we may use as input values for x. Because the denominator of a

fraction can’t be 0, therefore for the rational formulas

the zeroes of the denominator Q can’t be used as inputs.


PQ

DomainThe domain of a formula is the set of all the numbers that we may use as input values for x. Because the denominator of a



in other words, the domain of are all the numbers except where Q = 0.


PQ

PQ

Domain

b. Determine the domain of the formula

The domain of a formula is the set of all the numbers that we may use as input values for x. Because the denominator of a





x3 + x2 – 2xx2 + 4x + 3

PQ

PQ

Domain







x3 + x2 – 2xx2 + 4x + 3

x3 – 2x2 + 3xx2 + 4x + 3

Factor expression first.

= x(x + 2)(x – 1)(x + 3)(x + 1)

PQ

PQ

Domain







x3 + x2 – 2xx2 + 4x + 3

x3 – 2x2 + 3xx2 + 4x + 3


= x(x + 2)(x – 1)(x + 3)(x + 1)

Hence we can’t have

PQ

PQ

(x + 3)(x + 1) = 0

Domain







x3 + x2 – 2xx2 + 4x + 3

x3 – 2x2 + 3xx2 + 4x + 3


= x(x + 2)(x – 1)(x + 3)(x + 1)

Hence we can’t have

PQ

PQ

(x + 3)(x + 1) = 0so that the domain is the set of all the numbers except –1 and –3.

EvaluationIt is often easier to evaluate expressions in the factored form.


Evaluation

c. Evaluate if x = 7 using the factored form.

It is often easier to evaluate expressions in the factored form.


x3 + x2 – 2xx2 + 4x + 3

Evaluation




x3 + x2 – 2xx2 + 4x + 3

Using the factored form x(x + 2)(x – 1)(x + 3)(x + 1)

Plug in x = 7,

Evaluation




x3 + x2 – 2xx2 + 4x + 3


Plug in x = 7, we get 7(9)(6)(10)(8)

Evaluation




x3 + x2 – 2xx2 + 4x + 3


Plug in x = 7, we get 7(9)(6)(10)(8)

3

4

Evaluation




x3 + x2 – 2xx2 + 4x + 3


Plug in x = 7, we get 7(9)(6)(10)(8) =

3

418940

Evaluation




x3 + x2 – 2xx2 + 4x + 3


Plug in x = 7, we get 7(9)(6)(10)(8) =

3

418940

Signs

Evaluation




x3 + x2 – 2xx2 + 4x + 3


Plug in x = 7, we get 7(9)(6)(10)(8) =

3

418940

We use the factored form to determine the sign of an output.Signs

Evaluation




x3 + x2 – 2xx2 + 4x + 3


Plug in x = 7, we get 7(9)(6)(10)(8) =

3

418940


d. Determine the signs (positive or negative) ofif x = ½,– 5/2 using the factored form.

x3 + x2 – 2xx2 + 4x + 3

Evaluation




x3 + x2 – 2xx2 + 4x + 3


Plug in x = 7, we get 7(9)(6)(10)(8) =

3

418940



x3 + x2 – 2xx2 + 4x + 3

Plug in x = ½ into the above factored form and check the signs of each factor,

Evaluation




x3 + x2 – 2xx2 + 4x + 3


Plug in x = 7, we get 7(9)(6)(10)(8) =

3

418940



x3 + x2 – 2xx2 + 4x + 3

Plug in x = ½ into the above factored form and check the signs of each factor, we get +( + )( – )

(+)(+)

Evaluation




x3 + x2 – 2xx2 + 4x + 3


Plug in x = 7, we get 7(9)(6)(10)(8) =

3

418940



x3 + x2 – 2xx2 + 4x + 3

Plug in x = ½ into the above factored form and check the signs of each factor, we get = –, so it’s negative.+( + )( – )

(+)(+)

Evaluation




x3 + x2 – 2xx2 + 4x + 3


Plug in x = 7, we get 7(9)(6)(10)(8) =

3

418940



x3 + x2 – 2xx2 + 4x + 3

Plug in x = ½ into the above factored form and check the signs of each factor, we get

Plug in x = –5/2 into the factored form we get –( – )( – )(+)( – )

= –, so it’s negative.+( + )( – )(+)(+)

Evaluation




x3 + x2 – 2xx2 + 4x + 3


Plug in x = 7, we get 7(9)(6)(10)(8) =

3

418940



x3 + x2 – 2xx2 + 4x + 3

Plug in x = ½ into the above factored form and check the signs of each factor, we get = –, so it’s negative.+( + )( – )

(+)(+)Plug in x = –5/2 into the factored form we get –( – )( – )

(+)( – ) = +so that the output is positive.

PNQN = =

1

Rational ExpressionsCancellation Law: Common factor may be cancelled, i.e.

PNQN

PQ

PNQN = =

1


PNQN

PQ

A rational expression is reduced (simplified) if all common factors are cancelled.

PNQN = =

1


PNQN

PQ

A rational expression is reduced (simplified) if all common factors are cancelled.To reduce (simplify) a rational expression, put the expression in the factored form then cancel the common factors.

PNQN = =

1


PNQN

PQ

A rational expression is reduced (simplified) if all common factors are cancelled.To reduce (simplify) a rational expression, put the expression in the factored form then cancel the common factors.Example B. Reduce the following expressions.

(x – 2)(x + 3)(x + 3)(x + 2)a.

b. x2 – 3x + 10 x2 – 3

PNQN = =

1


PNQN

PQ


(x – 2)(x + 3)

It's already factored, proceed to cancel the common factor.(x + 3)(x + 2)

(x – 2)(x + 3)(x + 3)(x + 2)

a.

b. x2 – 3x + 10 x2 – 3

PNQN = =

1


PNQN

PQ


(x – 2)(x + 3)

1 = x – 2

x + 2


(x – 2)(x + 3)(x + 3)(x + 2) which is reduced.

a.

b. x2 – 3x + 10 x2 – 3

PNQN = =

1


PNQN

PQ


(x – 2)(x + 3)

1 = x – 2

x + 2



a.

b. x2 – 3x + 10 x2 – 3

This is in the factored form.

PNQN = =

1


PNQN

PQ


(x – 2)(x + 3)

1 = x – 2

x + 2



a.

b. x2 – 3x + 10 x2 – 3

This is in the factored form. There are no common factors so it’s already reduced.

Rational Expressionsc. x2 – 1

x2 – 3x+ 2


x2 – 3x+ 2

Factor then cancel


x2 – 3x+ 2x2 – 1

x2 – 3x+ 2= (x – 1)(x + 1) (x – 1)(x – 2)Factor then cancel


x2 – 3x+ 2x2 – 1

x2 – 3x+ 2= (x – 1)(x + 1) (x – 1)(x – 2)

1

Factor then cancel


x2 – 3x+ 2x2 – 1

x2 – 3x+ 2= (x – 1)(x + 1) (x – 1)(x – 2)

1

= x + 1 x – 2Factor then cancel


Only factors may be canceled.

c. x2 – 1 x2 – 3x+ 2

x2 – 1 x2 – 3x+ 2= (x – 1)(x + 1)

(x – 1)(x – 2)

1



Only factors may be canceled. It is wrong to cancel terms

c. x2 – 1 x2 – 3x+ 2

x2 – 1 x2 – 3x+ 2= (x – 1)(x + 1)

(x – 1)(x – 2)

1



Only factors may be canceled. It is wrong to cancel terms, i.e. P + Q P + R = P + Q

P + R

c. x2 – 1 x2 – 3x+ 2

x2 – 1 x2 – 3x+ 2= (x – 1)(x + 1)

(x – 1)(x – 2)

1




P + R

For example, x2 + 1 x2 – 2 = x2 + 1

x2 – 2

c. x2 – 1 x2 – 3x+ 2

x2 – 1 x2 – 3x+ 2= (x – 1)(x + 1)

(x – 1)(x – 2)

1




P + R

For example, x2 + 1 x2 – 2 = x2 + 1

x2 – 2

c. x2 – 1 x2 – 3x+ 2

x2 – 1 x2 – 3x+ 2= (x – 1)(x + 1)

(x – 1)(x – 2)

1


Cancellation of Opposite Factors



P + R

For example, x2 + 1 x2 – 2 = x2 + 1

x2 – 2

The opposite of a quantity x is the –x.

c. x2 – 1 x2 – 3x+ 2

x2 – 1 x2 – 3x+ 2= (x – 1)(x + 1)

(x – 1)(x – 2)

1





P + R

For example, x2 + 1 x2 – 2 = x2 + 1

x2 – 2


original: y -z x – y v – 4u – 2w

opposite:

c. x2 – 1 x2 – 3x+ 2

x2 – 1 x2 – 3x+ 2= (x – 1)(x + 1)

(x – 1)(x – 2)

1





P + R

For example, x2 + 1 x2 – 2 = x2 + 1

x2 – 2



opposite: -y

c. x2 – 1 x2 – 3x+ 2

x2 – 1 x2 – 3x+ 2= (x – 1)(x + 1)

(x – 1)(x – 2)

1





P + R

For example, x2 + 1 x2 – 2 = x2 + 1

x2 – 2



opposite: -y z

c. x2 – 1 x2 – 3x+ 2

x2 – 1 x2 – 3x+ 2= (x – 1)(x + 1)

(x – 1)(x – 2)

1





P + R

For example, x2 + 1 x2 – 2 = x2 + 1

x2 – 2



opposite: -y z -x + y or y – x

c. x2 – 1 x2 – 3x+ 2

x2 – 1 x2 – 3x+ 2= (x – 1)(x + 1)

(x – 1)(x – 2)

1





P + R

For example, x2 + 1 x2 – 2 = x2 + 1

x2 – 2



opposite: -y z -x + y or y – x -v + 4u + 2w or …

c. x2 – 1 x2 – 3x+ 2

x2 – 1 x2 – 3x+ 2= (x – 1)(x + 1)

(x – 1)(x – 2)

1





P + R

For example, x2 + 1 x2 – 2 = x2 + 1

x2 – 2Cancellation of Opposite Factors


While identical factors cancel to be 1, opposite factors cancel to be –1,



c. x2 – 1 x2 – 3x+ 2

x2 – 1 x2 – 3x+ 2= (x – 1)(x + 1)

(x – 1)(x – 2)

1




P + R

For example, x2 + 1 x2 – 2 = x2 + 1

x2 – 2


While identical factors cancel to be 1, opposite factors cancel to be –1, in symbol,



x –x = –

1.

c. x2 – 1 x2 – 3x+ 2

x2 – 1 x2 – 3x+ 2= (x – 1)(x + 1)

(x – 1)(x – 2)

1



Example C.

2y–2y

a.


Example C.

2y–2y =

-1–1a.


b(x – y)a(y – x)

Example C.

2y–2y =

-1–1a.

b.


b(x – y)a(y – x)

Example C.

2y–2y =

-1–1

-1

a.

b.


b(x – y)a(y – x)

Example C.

2y–2y =

-1–1

-1= –b

a

a.

b.


b(x – y)a(y – x)

Example C.

2y–2y =

-1–1

-1= –b

a

a.

b.


(x – 2y)(x + 3y) (x – 3y)(2y – x)

c.

b(x – y)a(y – x)

Example C.

2y–2y =

-1–1

-1= –b

a

a.

b.


(x – 2y)(x + 3y) (x – 3y)(2y – x)

c.-1

b(x – y)a(y – x)

Example C.

2y–2y =

-1–1

-1= –b

a

a.

b.


(x – 2y)(x + 3y) (x – 3y)(2y – x)

c.-1

= –(x + 3y) (x – 3y)

b(x – y)a(y – x)

Example C.

2y–2y =

-1–1

-1= –b

a

a.

b.


(x – 2y)(x + 3y) (x – 3y)(2y – x)

c.-1

= –(x + 3y) (x – 3y) or –x – 3y

x – 3y

b(x – y)a(y – x)

Example C.

2y–2y =

-1–1

-1= –b

a

a.

b.


(x – 2y)(x + 3y) (x – 3y)(2y – x)

c.-1

= –(x + 3y) (x – 3y) or –x – 3y

x – 3y

d. 4 – x2

x2 – x – 2

b(x – y)a(y – x)

Example C.

2y–2y =

-1–1

-1= –b

a

a.

b.


(x – 2y)(x + 3y) (x – 3y)(2y – x)

c.-1

= –(x + 3y) (x – 3y) or –x – 3y

x – 3y

d. 4 – x2

x2 – x – 2 = (2 – x)(2 + x)(x + 1)(x – 2)

b(x – y)a(y – x)

Example C.

2y–2y =

-1–1

-1= –b

a

a.

b.


(x – 2y)(x + 3y) (x – 3y)(2y – x)

c.-1

= –(x + 3y) (x – 3y) or –x – 3y

x – 3y

d. 4 – x2

x2 – x – 2 = (2 – x)(2 + x)(x + 1)(x – 2)

-1

b(x – y)a(y – x)

Example C.

2y–2y =

-1–1

-1= –b

a

a.

b.


(x – 2y)(x + 3y) (x – 3y)(2y – x)

c.-1

= –(x + 3y) (x – 3y) or –x – 3y

x – 3y

d. 4 – x2

x2 – x – 2 = (2 – x)(2 + x)(x + 1)(x – 2)

-1= –(2 + x)

x + 1 or –2 – x x + 1

b(x – y)a(y – x)

Example C.

2y–2y =

-1–1

-1= –b

a

a.

b.


(x – 2y)(x + 3y) (x – 3y)(2y – x)

c.-1

= –(x + 3y) (x – 3y) or –x – 3y

x – 3y

d. 4 – x2

x2 – x – 2 = (2 – x)(2 + x)(x + 1)(x – 2)

-1= –(2 + x)

x + 1 or –2 – x x + 1

In the case of polynomials in one variable x, if the highestdegree term is negative, we may factor out the negative sign then factor the expressions.

b(x – y)a(y – x)

Example C.

2y–2y =

-1–1

-1= –b

a

a.

b.


(x – 2y)(x + 3y) (x – 3y)(2y – x)

c.-1

= –(x + 3y) (x – 3y) or –x – 3y

x – 3y

d. 4 – x2

x2 – x – 2 = (2 – x)(2 + x)(x + 1)(x – 2)

-1= –(2 + x)

x + 1 or –2 – x x + 1

–x2 + 4–x2 + x + 2

Example D. Pull out the “–” first then reduce.


b(x – y)a(y – x)

Example C.

2y–2y =

-1–1

-1= –b

a

a.

b.


(x – 2y)(x + 3y) (x – 3y)(2y – x)

c.-1

= –(x + 3y) (x – 3y) or –x – 3y

x – 3y

d. 4 – x2

x2 – x – 2 = (2 – x)(2 + x)(x + 1)(x – 2)

-1= –(2 + x)

x + 1 or –2 – x x + 1


–x2 + 4–x2 + x + 2 =

Example D. Pull out the “–” first then reduce. –(x2 – 4)

–(x2 – x – 2)

b(x – y)a(y – x)

Example C.

2y–2y =

-1–1

-1= –b

a

a.

b.


(x – 2y)(x + 3y) (x – 3y)(2y – x)

c.-1

= –(x + 3y) (x – 3y) or –x – 3y

x – 3y

d. 4 – x2

x2 – x – 2 = (2 – x)(2 + x)(x + 1)(x – 2)

-1= –(2 + x)

x + 1 or –2 – x x + 1

–x2 + 4–x2 + x + 2 = (x – 2)(x + 2)

(x + 1)(x – 2)=


–(x2 – x – 2)


b(x – y)a(y – x)

Example C.

2y–2y =

-1–1

-1= –b

a

a.

b.


(x – 2y)(x + 3y) (x – 3y)(2y – x)

c.-1

= –(x + 3y) (x – 3y) or –x – 3y

x – 3y

d. 4 – x2

x2 – x – 2 = (2 – x)(2 + x)(x + 1)(x – 2)

-1= –(2 + x)

x + 1 or –2 – x x + 1

–x2 + 4–x2 + x + 2 = (x – 2)(x + 2)

(x + 1)(x – 2)=


–(x2 – x – 2) = x + 2 x + 1


Rational ExpressionsTo summarize, a rational expression is reduced (simplified) if all common factors are cancelled.Following are the steps for reducing a rational expression.1. Factor the top and bottom completely. (If present, factor the “ – ” from the leading term)2. Cancel the common factors: -cancel identical factors to be 1 -cancel opposite factors to be –1

Ex. A. Write the following expressions in factored form. List all the distinct factors of the numerator and the denominator of each expression.

1.


2x + 3 x + 3 2. 4x + 6

2x + 6 3. x2 – 4 2x + 4

4. x2 + 4x2 + 4x 5. x2 – 2x – 3

x2 + 4x 6. x3 – 2x2 – 8xx2 + 2x – 3

7. Find the zeroes and list the domain of x2 – 2x – 3x2 + 4x

8. Use the factored form to evaluate x2 – 2x – 3x2 + 4x

with x = 7, ½, – ½, 1/3.

9. Determine the signs of the outputs of x2 – 2x – 3x2 + 4x

with x = 4, –2, 1/7, 1.23.

For problems 10, 11, and 12, answer the same questions

as problems 7, 8 and 9 with the formula .x3 – 2x2 – 8xx2 + 2x – 3

Ex. B. Reduce the following expressions. If it’s already reduced, state this. Make sure you do not cancel any terms and make sure that you look for the opposite cancellation.

13.


2x + 3 x + 3

20. 4x + 6 2x + 3

22. 23. 24.

21.

3x – 12x – 4

12 – 3xx – 4

4x + 6 –2x – 3

3x + 12x – 4

25. 4x – 6 –2x – 3

14. x + 3 x – 3 15. x + 3

–x – 3

16. x + 3x – 3

17. x – 3 3 – x 18. 2x – 1

1 + 2x

19. 2x – 1 1 – 2x

26. (2x – y)(x – 2y)(2y + x)(y – 2x) 27. (3y + x)(3x –y)

(y – 3x)(–x – 3y)

28. (2u + v – w)(2v – u – 2w)(u – 2v + 2w)(–2u – v – w) 29. (a + 4b – c)(a – b – c)

(c – a – 4b)(a + b + c)

30.


37.

x2 – 1x2 + 2x – 3

36. 38. x – x2

39. x2 – 3x – 4

31. 32.

33. 34. 35.

40. 41. x3 – 16x

x2 + 4 2x + 4

x2 – 4x + 4

x2 – 4x2– 2x

x2 – 9x2 + 4x + 3

x2 – 4 2x + 4

x2 + 3x + 2x2 – x – 2 x2 + x – 2

x2 – x – 6x2 – 5x + 6

x2 – x – 2x2 + x – 2

x2 – 5x – 6

x2 + 5x – 6x2 + 5x + 6

x3 – 8x2 – 20x

46.45. 47. 9 – x2

42. 43. 44. x2 – 2x9 – x2

x2 + 4x + 3– x2 – x + 2

x3 – x2 – 6x–1 + x2

–x2 + x + 2x2 – x – 2

– x2 + 5x – 61 – x2

x2 + 5x – 6

49.48. 50.xy – 2y + x2 – 2xx2 – y2 x3 – 100x

x2 – 4xy + x – 4yx2 – 3xy – 4y2

Ex. C. Reduce the following expressions. If it’s already reduced, state this. Make sure you do not cancel any terms and make sure that you look for the opposite cancellation.

Education

12 rational expressions