1 2 rational expressions

Rational Expressions
Frank Ma © 2011

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,

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

For example,
Rational (fractional) expressions are expressions of the
form , where P and Q are polynomials.
P
Q

For example,
P
Q
P
All polynomials are rational expressions by viewing P as .
1

For example,
P
Q
P
1
x – 2
x2 – 2 x + 1 ,

For example,
P
Q
P
1
x – 2
x(x – 2)
are rational expressions.
x2 – 2 x + 1 ,
(x + 1) (2x + 1)

For example,
P
Q
P
1
x – 2
x(x – 2)
x2 – 2 x + 1 ,
(x + 1) (2x + 1)
x – 2
is not a rational expression because the
 2 x + 1
denominator is not a polynomial.

For example,
P
Q
P
1
x – 2
x(x – 2)
x2 – 2 x + 1 ,
(x + 1) (2x + 1)
x – 2
is not a rational expression because the
 2 x + 1
denominator is not a polynomial.
Rational expressions are expressions that describe calculation procedures that involve division (of polynomials).

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

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

x2 – 4
x2+ 2x + 1
(x + 2)(x – 2)
In the factored form,it’s
(x + 1)(x + 1) .

x2 – 4
x2+ 2x + 1
(x + 2)(x – 2)
(x + 1)(x + 1) .
x – 2
The expression
is in both forms.
x2 + 1
Example A. Put the following expressions in the factored form.
x2 – 3x – 10
a.
x2 – 3x
x2 – 3x + 10
b.
x2 – 3

x2 – 4
x2+ 2x + 1
(x + 2)(x – 2)
(x + 1)(x + 1) .
x – 2
The expression
is in both forms.
x2 + 1
(x – 5)(x + 2)
x2 – 3x – 10
a.
=
x(x – 3)
x2 – 3x
x2 – 3x + 10
b.
x2 – 3

x2 – 4
x2+ 2x + 1
(x + 2)(x – 2)
(x + 1)(x + 1) .
x – 2
The expression
is in both forms.
x2 + 1
(x – 5)(x + 2)
x2 – 3x – 10
a.
=
x(x – 3)
x2 – 3x
x2 – 3x + 10
b.
is in the factored form
x2 – 3

x2 – 4
x2+ 2x + 1
(x + 2)(x – 2)
(x + 1)(x + 1) .
x – 2
The expression
is in both forms.
x2 + 1
(x – 5)(x + 2)
x2 – 3x – 10
a.
=
x(x – 3)
x2 – 3x
x2 – 3x + 10
b.
is in the factored form
x2 – 3
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

1. solve equations
2. determine the domain of rational expressions

1. solve equations
2. evaluate given inputs

1. solve equations
3. determine of the signs of the outputs

1. solve equations
Solutions of Equations

1. solve equations
P
The solutions of the equation of the form = 0 are the
zeroes of the numerator, so they are the solutions of P = 0.
Q

1. solve equations
P
Q
x3+ x2 – 2x
Example A. a. Write in the factored form and
solve the equation
x2 + 4x + 3
x3+ x2 – 2x
= 0
x2 + 4x + 3

1. solve equations
P
Q
x3+ x2 – 2x
solve the equation
x2 + 4x + 3
x3+ x2 – 2x
= 0
x2 + 4x + 3
x3 – 2x2 + 3x
=
Factor, we get
x2 + 4x + 3

1. solve equations
P
Q
x3+ x2 – 2x
solve the equation
x2 + 4x + 3
x3+ x2 – 2x
= 0
x2 + 4x + 3
x3 – 2x2 + 3x
x(x + 2)(x – 1)
=
Factor, we get
x2 + 4x + 3
(x + 3)(x + 1)

1. solve equations
P
Q
x3+ x2 – 2x
solve the equation
x2 + 4x + 3
x3+ x2 – 2x
= 0
x2 + 4x + 3
x3 – 2x2 + 3x
x(x + 2)(x – 1)
=
Factor, we get
x2 + 4x + 3
(x + 3)(x + 1)
x3+ x2 – 2x
Hence for
it must be that x(x + 2)(x – 1) = 0
= 0,
x2 + 4x + 3

1. solve equations
P
Q
x3+ x2 – 2x
solve the equation
x2 + 4x + 3
x3+ x2 – 2x
= 0
x2 + 4x + 3
x3 – 2x2 + 3x
x(x + 2)(x – 1)
=
Factor, we get
x2 + 4x + 3
(x + 3)(x + 1)
x3+ x2 – 2x
Hence for
it must be that x(x + 2)(x – 1) = 0
= 0,
x2 + 4x + 3
or that x = 0, –2, 1.

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

Domain
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
fraction can’t be 0, therefore for the rational formulas
the zeroes of the denominator Q can’t be used as inputs.
P
Q

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

Domain
P
Q
P
Q
x3+ x2 – 2x
b. Determine the domain of the formula
x2 + 4x + 3

Domain
P
Q
P
Q
x3+ x2 – 2x
x2 + 4x + 3
Factor expression first.
x3 – 2x2 + 3x
x(x + 2)(x – 1)
=
x2 + 4x + 3
(x + 3)(x + 1)

Domain
P
Q
P
Q
x3+ x2 – 2x
x2 + 4x + 3
x3 – 2x2 + 3x
x(x + 2)(x – 1)
=
x2 + 4x + 3
(x + 3)(x + 1)
Hence we can’t have
(x + 3)(x + 1) = 0

Domain
P
Q
P
Q
x3+ x2 – 2x
x2 + 4x + 3
x3 – 2x2 + 3x
x(x + 2)(x – 1)
=
x2 + 4x + 3
(x + 3)(x + 1)
Hence we can’t have
(x + 3)(x + 1) = 0
so that the domain is the set of all the numbers except
–1 and –3.

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

Evaluation
x3+ x2 – 2x
c. Evaluate if x = 7 using the factored form.
x2 + 4x + 3

Evaluation
x3+ x2 – 2x
x2 + 4x + 3
x(x + 2)(x – 1)
Using the factored form
(x + 3)(x + 1)
plug in x = 7,

Evaluation
x3+ x2 – 2x
x2 + 4x + 3
x(x + 2)(x – 1)
(x + 3)(x + 1)
7(9)(6)
plug in x = 7, we get
(10)(8)

Evaluation
x3+ x2 – 2x
x2 + 4x + 3
x(x + 2)(x – 1)
(x + 3)(x + 1)
3
7(9)(6)
(10)(8)
4

Evaluation
x3+ x2 – 2x
x2 + 4x + 3
x(x + 2)(x – 1)
(x + 3)(x + 1)
3
7(9)(6)
189
=
(10)(8)
40
4

Evaluation
x3+ x2 – 2x
x2 + 4x + 3
x(x + 2)(x – 1)
(x + 3)(x + 1)
3
7(9)(6)
189
=
(10)(8)
40
4
Signs

Evaluation
x3+ x2 – 2x
x2 + 4x + 3
x(x + 2)(x – 1)
(x + 3)(x + 1)
3
7(9)(6)
189
=
(10)(8)
40
4
Signs
We use the factored form to determine the sign of the outputs.

Evaluation
x3+ x2 – 2x
x2 + 4x + 3
x(x + 2)(x – 1)
(x + 3)(x + 1)
3
7(9)(6)
189
=
(10)(8)
40
4
Signs
x3+ x2 – 2x
d. Determine the signs (positive or negative) of
if x = ½,– 5/2 using the factored form.
x2 + 4x + 3

Evaluation
x3+ x2 – 2x
x2 + 4x + 3
x(x + 2)(x – 1)
(x + 3)(x + 1)
3
7(9)(6)
189
=
(10)(8)
40
4
Signs
x3+ x2 – 2x
x2 + 4x + 3
Plug in x = ½ into the above factored form and check the signs of each factor,

Evaluation
x3+ x2 – 2x
x2 + 4x + 3
x(x + 2)(x – 1)
(x + 3)(x + 1)
3
7(9)(6)
189
=
(10)(8)
40
4
Signs
x3+ x2 – 2x
x2 + 4x + 3
Plug in x = ½ into the above factored form and check the signs of each factor, we get
+( + )( – )
(+)(+)

Evaluation
x3+ x2 – 2x
x2 + 4x + 3
x(x + 2)(x – 1)
(x + 3)(x + 1)
3
7(9)(6)
189
=
(10)(8)
40
4
Signs
x3+ x2 – 2x
x2 + 4x + 3
+( + )( – )
= –, so it’s negative.
(+)(+)

Evaluation
x3+ x2 – 2x
x2 + 4x + 3
x(x + 2)(x – 1)
(x + 3)(x + 1)
3
7(9)(6)
189
=
(10)(8)
40
4
Signs
x3+ x2 – 2x
x2 + 4x + 3
+( + )( – )
(+)(+)
–( – )( – )
Plug in x = –5/2 into the factored form we get
(+)( – )

Evaluation
x3+ x2 – 2x
x2 + 4x + 3
x(x + 2)(x – 1)
(x + 3)(x + 1)
3
7(9)(6)
189
=
(10)(8)
40
4
Signs
x3+ x2 – 2x
x2 + 4x + 3
+( + )( – )
(+)(+)
–( – )( – )
Plug in x = –5/2 into the factored form we get
= +
(+)( – )
so that the output is positive.

Cancellation Law: Common factor may be canceled, i.e.
1
PN
PN
P
=
=
QN
QN
Q

1
PN
PN
P
=
=
QN
QN
Q
A rational expression is reduced (simplified)if all common factors are canceled.

1
PN
PN
P
=
=
QN
QN
Q
To reduce (simplify) a rational expression, put the expression in the factored form then cancel the common factors.

1
PN
PN
P
=
=
QN
QN
Q
Example B. Reduce the following expressions.
(x – 2)(x + 3)
a.
(x + 3)(x + 2)
x2 – 3x + 10
b.
x2 – 3

1
PN
PN
P
=
=
QN
QN
Q
(x – 2)(x + 3)
a.
(x + 3)(x + 2)
It's already factored, proceed to cancel the common factor.
(x – 2)(x + 3)
(x + 3)(x + 2)
x2 – 3x + 10
b.
x2 – 3

1
PN
PN
P
=
=
QN
QN
Q
(x – 2)(x + 3)
a.
(x + 3)(x + 2)
1
x – 2
(x – 2)(x + 3)
=
which is reduced.
x + 2
(x + 3)(x + 2)
x2 – 3x + 10
b.
x2 – 3

1
PN
PN
P
=
=
QN
QN
Q
(x – 2)(x + 3)
a.
(x + 3)(x + 2)
1
x – 2
(x – 2)(x + 3)
=
which is reduced.
x + 2
(x + 3)(x + 2)
x2 – 3x + 10
b.
x2 – 3
This is in the factored form.

1
PN
PN
P
=
=
QN
QN
Q
(x – 2)(x + 3)
a.
(x + 3)(x + 2)
1
x – 2
(x – 2)(x + 3)
=
which is reduced.
x + 2
(x + 3)(x + 2)
x2 – 3x + 10
b.
x2 – 3
This is in the factored form. There are no common factors so it’s already reduced.

x2 – 1
c.
x2 – 3x+ 2

x2 – 1
c.
x2 – 3x+ 2
Factor then cancel

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

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

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

x2 – 1
c.
x2 – 3x+ 2
1
x2 – 1
(x – 1)(x + 1)
x + 1
=
=
Factor then cancel
x2 – 3x+ 2
(x – 1)(x – 2)
x – 2
Only factors may be canceled.

x2 – 1
c.
x2 – 3x+ 2
1
x2 – 1
(x – 1)(x + 1)
x + 1
=
=
Factor then cancel
x2 – 3x+ 2
(x – 1)(x – 2)
x – 2
Only factors may be canceled. It is wrong to cancel terms

x2 – 1
c.
x2 – 3x+ 2
1
x2 – 1
(x – 1)(x + 1)
x + 1
=
=
Factor then cancel
x2 – 3x+ 2
(x – 1)(x – 2)
x – 2
Only factors may be canceled. It is wrong to cancel terms, i.e.
P + Q
P + Q
=
P + R
P + R

x2 – 1
c.
x2 – 3x+ 2
1
x2 – 1
(x – 1)(x + 1)
x + 1
=
=
Factor then cancel
x2 – 3x+ 2
(x – 1)(x – 2)
x – 2
P + Q
P + Q
=
P + R
P + R
x2 + 1
x2 + 1
For example,
=
x2 – 2
x2 – 2

x2 – 1
c.
x2 – 3x+ 2
1
x2 – 1
(x – 1)(x + 1)
x + 1
=
=
Factor then cancel
x2 – 3x+ 2
(x – 1)(x – 2)
x – 2
P + Q
P + Q
=
P + R
P + R
x2 + 1
x2 + 1
For example,
=
x2 – 2
x2 – 2
Cancellation of Opposite Factor

x2 – 1
c.
x2 – 3x+ 2
1
x2 – 1
(x – 1)(x + 1)
x + 1
=
=
Factor then cancel
x2 – 3x+ 2
(x – 1)(x – 2)
x – 2
P + Q
P + Q
=
P + R
P + R
x2 + 1
x2 + 1
For example,
=
x2 – 2
x2 – 2
The opposite of a quantity x is the –x.

x2 – 1
c.
x2 – 3x+ 2
1
x2 – 1
(x – 1)(x + 1)
x + 1
=
=
Factor then cancel
x2 – 3x+ 2
(x – 1)(x – 2)
x – 2
P + Q
P + Q
=
P + R
P + R
x2 + 1
x2 + 1
For example,
=
x2 – 2
x2 – 2
original: y -z x – y v – 4u – 2w
opposite:

x2 – 1
c.
x2 – 3x+ 2
1
x2 – 1
(x – 1)(x + 1)
x + 1
=
=
Factor then cancel
x2 – 3x+ 2
(x – 1)(x – 2)
x – 2
P + Q
P + Q
=
P + R
P + R
x2 + 1
x2 + 1
For example,
=
x2 – 2
x2 – 2
opposite: -y

x2 – 1
c.
x2 – 3x+ 2
1
x2 – 1
(x – 1)(x + 1)
x + 1
=
=
Factor then cancel
x2 – 3x+ 2
(x – 1)(x – 2)
x – 2
P + Q
P + Q
=
P + R
P + R
x2 + 1
x2 + 1
For example,
=
x2 – 2
x2 – 2
opposite: -y z

x2 – 1
c.
x2 – 3x+ 2
1
x2 – 1
(x – 1)(x + 1)
x + 1
=
=
Factor then cancel
x2 – 3x+ 2
(x – 1)(x – 2)
x – 2
P + Q
P + Q
=
P + R
P + R
x2 + 1
x2 + 1
For example,
=
x2 – 2
x2 – 2
opposite: -y z -x + y or y – x

x2 – 1
c.
x2 – 3x+ 2
1
x2 – 1
(x – 1)(x + 1)
x + 1
=
=
Factor then cancel
x2 – 3x+ 2
(x – 1)(x – 2)
x – 2
P + Q
P + Q
=
P + R
P + R
x2 + 1
x2 + 1
For example,
=
x2 – 2
x2 – 2
opposite: -y z -x + y or y – x -v + 4u + 2w or …

x2 – 1
c.
x2 – 3x+ 2
1
x2 – 1
(x – 1)(x + 1)
x + 1
=
=
Factor then cancel
x2 – 3x+ 2
(x – 1)(x – 2)
x – 2
P + Q
P + Q
=
P + R
P + R
x2 + 1
x2 + 1
For example,
=
x2 – 2
x2 – 2
While identical factors cancel to be 1, opposite factors
cancel to be –1,

x2 – 1
c.
x2 – 3x+ 2
1
x2 – 1
(x – 1)(x + 1)
x + 1
=
=
Factor then cancel
x2 – 3x+ 2
(x – 1)(x – 2)
x – 2
P + Q
P + Q
=
P + R
P + R
x2 + 1
x2 + 1
For example,
=
x2 – 2
x2 – 2
While identical factors cancel to be 1, opposite factors
cancel to be –1, in symbol,
x
= –1.
–x

Example C.
2y
a.
–2y

Example C.
-1
2y
a.
=
–1
–2y

Example C.
-1
2y
a.
=
–1
–2y
b(x – y)
b.
a(y – x)

Example C.
-1
2y
a.
=
–1
–2y
-1
b(x – y)
b.
a(y – x)

Example C.
-1
2y
a.
=
–1
–2y
-1
b(x – y)
–b
=
b.
a(y – x)
a

Example C.
-1
2y
a.
=
–1
–2y
-1
b(x – y)
–b
=
b.
a(y – x)
a
(x – 2y)(x + 3y)
c.
(x – 3y)(2y – x)

Example C.
-1
2y
a.
=
–1
–2y
-1
b(x – y)
–b
=
b.
a(y – x)
a
-1
(x – 2y)(x + 3y)
c.
(x – 3y)(2y – x)

Example C.
-1
2y
a.
=
–1
–2y
-1
b(x – y)
–b
=
b.
a(y – x)
a
-1
(x – 2y)(x + 3y)
–(x + 3y)
c.
=
(x – 3y)(2y – x)
(x – 3y)

Example C.
-1
2y
a.
=
–1
–2y
-1
b(x – y)
–b
=
b.
a(y – x)
a
-1
(x – 2y)(x + 3y)
–(x + 3y)
–x – 3y
c.
=
or
(x – 3y)(2y – x)
(x – 3y)
x – 3y

Example C.
-1
2y
a.
=
–1
–2y
-1
b(x – y)
–b
=
b.
a(y – x)
a
-1
(x – 2y)(x + 3y)
–(x + 3y)
–x – 3y
c.
=
or
(x – 3y)(2y – x)
(x – 3y)
x – 3y
4 – x2
d.
x2 – x – 2

Example C.
-1
2y
a.
=
–1
–2y
-1
b(x – y)
–b
=
b.
a(y – x)
a
-1
(x – 2y)(x + 3y)
–(x + 3y)
–x – 3y
c.
=
or
(x – 3y)(2y – x)
(x – 3y)
x – 3y
4 – x2
(2 – x)(2 + x)
d.
=
x2 – x – 2
(x+ 1)(x – 2)

Example C.
-1
2y
a.
=
–1
–2y
-1
b(x – y)
–b
=
b.
a(y – x)
a
-1
(x – 2y)(x + 3y)
–(x + 3y)
–x – 3y
c.
=
or
(x – 3y)(2y – x)
(x – 3y)
x – 3y
-1
4 – x2
(2 – x)(2 + x)
d.
=
x2 – x – 2
(x+ 1)(x – 2)

Example C.
-1
2y
a.
=
–1
–2y
-1
b(x – y)
–b
=
b.
a(y – x)
a
-1
(x – 2y)(x + 3y)
–(x + 3y)
–x – 3y
c.
=
or
(x – 3y)(2y – x)
(x – 3y)
x – 3y
-1
4 – x2
(2 – x)(2 + x)
–(2 + x)
–2 – x
d.
=
=
or
x2 – x – 2
(x+ 1)(x – 2)
x + 1
x + 1

Example C.
-1
2y
a.
=
–1
–2y
-1
b(x – y)
–b
=
b.
a(y – x)
a
-1
(x – 2y)(x + 3y)
–(x + 3y)
–x – 3y
c.
=
or
(x – 3y)(2y – x)
(x – 3y)
x – 3y
-1
4 – x2
(2 – x)(2 + x)
–(2 + x)
–2 – x
d.
=
=
or
x2 – x – 2
(x+ 1)(x – 2)
x + 1
x + 1
In the case of polynomials in one variable x, if the highest
degree term is negative, we may factor out the negative sign
then factor the expressions.

Example C.
-1
2y
a.
=
–1
–2y
-1
b(x – y)
–b
=
b.
a(y – x)
a
-1
(x – 2y)(x + 3y)
–(x + 3y)
–x – 3y
c.
=
or
(x – 3y)(2y – x)
(x – 3y)
x – 3y
-1
4 – x2
(2 – x)(2 + x)
–(2 + x)
–2 – x
d.
=
=
or
x2 – x – 2
(x+ 1)(x – 2)
x + 1
x + 1
Example D. Pull out the “–” first then reduce.
–x2 + 4
–x2 + x + 2

Example C.
-1
2y
a.
=
–1
–2y
-1
b(x – y)
–b
=
b.
a(y – x)
a
-1
(x – 2y)(x + 3y)
–(x + 3y)
–x – 3y
c.
=
or
(x – 3y)(2y – x)
(x – 3y)
x – 3y
-1
4 – x2
(2 – x)(2 + x)
–(2 + x)
–2 – x
d.
=
=
or
x2 – x – 2
(x+ 1)(x – 2)
x + 1
x + 1
–x2 + 4
–(x2 – 4)
=
–x2 + x + 2
–(x2 – x – 2)

Example C.
-1
2y
a.
=
–1
–2y
-1
b(x – y)
–b
=
b.
a(y – x)
a
-1
(x – 2y)(x + 3y)
–(x + 3y)
–x – 3y
c.
=
or
(x – 3y)(2y – x)
(x – 3y)
x – 3y
-1
4 – x2
(2 – x)(2 + x)
–(2 + x)
–2 – x
d.
=
=
or
x2 – x – 2
(x+ 1)(x – 2)
x + 1
x + 1
–x2 + 4
(x – 2)(x + 2)
–(x2 – 4)
=
=
–x2 + x + 2
(x+ 1)(x – 2)
–(x2 – x – 2)

Example C.
-1
2y
a.
=
–1
–2y
-1
b(x – y)
–b
=
b.
a(y – x)
a
-1
(x – 2y)(x + 3y)
–(x + 3y)
–x – 3y
c.
=
or
(x – 3y)(2y – x)
(x – 3y)
x – 3y
-1
4 – x2
(2 – x)(2 + x)
–(2 + x)
–2 – x
d.
=
=
or
x2 – x – 2
(x+ 1)(x – 2)
x + 1
x + 1
–x2 + 4
(x – 2)(x + 2)
–(x2 – 4)
x + 2
=
=
=
–x2 + x + 2
(x+ 1)(x – 2)
–(x2 – x – 2)
x + 1

To summarize, a rational expression is reduced (simplified)
if all common factors are canceled.
Following are the steps for reducing a rational expression.
Factor the top and bottom completely.
(If present, factor the “ – ” from the leading term)
Cancel the common factors:
-cancel identical factor 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.
2x + 3
4x + 6
x2 – 4
1.
2.
3.
x + 3
2x + 6
2x + 4
x2 + 4
x2 – 2x – 3
x3 – 2x2– 8x
4.
5.
6.
x2 + 4x
x2 + 4x
x2+ 2x – 3
x2 – 2x – 3
7. Find the zeroes and list the domain of
x2 + 4x
x2 – 2x – 3
8. Use the factored form to evaluate
x2 + 4x
with x = 7, ½, – ½, 1/3.
x2 – 2x – 3
9. Determine the signs of the outputs of
x2+ 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– 8x
x2 + 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.
x + 3
x + 3
2x + 3
14.
15.
13.
x – 3
–x – 3
x + 3
x – 3
2x – 1
x + 3
17.
18.
16.
3 – x
1 + 2x
x – 3
4x + 6
4x + 6
2x – 1
21.
20.
19.
–2x – 3
2x + 3
1 – 2x
3x + 12
3x – 12
12 – 3x
24.
22.
23.
x – 4
x – 4
x – 4
4x – 6
(3y + x)(3x –y)
(2x – y)(x – 2y)
25.
27.
26.
–2x – 3
(y – 3x)(–x – 3y)
(2y + x)(y – 2x)
(2u + v – w)(2v – u – 2w)
(a + 4b – c)(a – b – c)
28.
29.
(u – 2v + 2w)(–2u – v – w)
(c – a – 4b)(a + b + c)

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.
x2+ 4
x2 – 4
x2– 4
31.
32.
2x + 4
30.
x2– 2x
2x + 4
x2+ x – 2
x2– 9
x2– x – 2
34.
35.
33.
x2– 4x + 4
x2 + 4x + 3
x2 + 3x + 2
x2– x – 6
x2– x – 2
x2– 5x – 6
36.
38.
37.
x2– 5x + 6
x2+ x – 2
x – x2
x2– 3x – 4
x2 + 5x – 6
x2+ 2x – 3
41.
39.
40.
x3 – 16x
x2 + 5x + 6
x2– 1
– x2 – x + 2
x2 + 5x – 6
9 – x2
43.
44.
42.
x2– 2x
1 – x2
x2 + 4x + 3
x3– x2– 6x
– x2 + 5x – 6
–x2 + x + 2
45.
47.
46.
–1 + x2
x2 – x – 2
9 – x2
xy – 2y + x2 – 2x
x3– 8x2– 20x
x2 – 4xy + x – 4y
48.
50.
49.
x2 – y2
x3 – 100x
x2 – 3xy – 4y2

1 2 rational expressions

by math123b on Aug 28, 2011

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1 2 rational expressions — Presentation Transcript