1 4 multiplication and division of rational expressions

Multiplication and Division of Rational Expressions
Frank Ma © 2011

Multiplication Rule for Rational Expressions
A
C
AC
*
=
B
D
BD

A
C
AC
*
=
B
D
BD
In most problems, we reduce the product by factoring the top and the bottom, then cancel.

A
C
AC
*
=
B
D
BD
Example A. Simplify
10x
y2
a.
*
y3z
5x3

A
C
AC
*
=
B
D
BD
Example A. Simplify
10x
y2
10xy2
a.
=
*
y3z
5x3
5x3y3z

A
C
AC
*
=
B
D
BD
Example A. Simplify
10x
y2
10xy2
2
a.
=
=
*
y3z
5x3
5x3y3z
x2yz

A
C
AC
*
=
B
D
BD
Example A. Simplify
10x
y2
10xy2
2
a.
=
=
*
y3z
5x3
5x3y3z
x2yz
(x2 + 2x – 3 )
(x2 – 4 )
b.
*
(x – 2)
(x2– x )

A
C
AC
*
=
B
D
BD
Example A. Simplify
10x
y2
10xy2
2
a.
=
=
*
y3z
5x3
5x3y3z
x2yz
(x2 + 2x – 3 )
(x2 – 4 )
b.
*
(x – 2)
(x2– x )
(x + 3)(x – 1 )
=

A
C
AC
*
=
B
D
BD
Example A. Simplify
10x
y2
10xy2
2
a.
=
=
*
y3z
5x3
5x3y3z
x2yz
(x2 + 2x – 3 )
(x2 – 4 )
b.
*
(x – 2)
(x2– x )
(x + 3)(x – 1 )
(x + 2 )(x – 2)
=

A
C
AC
*
=
B
D
BD
Example A. Simplify
10x
y2
10xy2
2
a.
=
=
*
y3z
5x3
5x3y3z
x2yz
(x2 + 2x – 3 )
(x2 – 4 )
b.
*
(x – 2)
(x2– x )
(x + 3)(x – 1 )
(x + 2 )(x – 2)
=
(x – 2)

A
C
AC
*
=
B
D
BD
Example A. Simplify
10x
y2
10xy2
2
a.
=
=
*
y3z
5x3
5x3y3z
x2yz
(x2 + 2x – 3 )
(x2 – 4 )
b.
*
(x – 2)
(x2– x )
(x + 3)(x – 1 )
(x + 2 )(x – 2)
=
(x – 2)
x(x – 1 )

A
C
AC
*
=
B
D
BD
Example A. Simplify
10x
y2
10xy2
2
a.
=
=
*
y3z
5x3
5x3y3z
x2yz
(x2 + 2x – 3 )
(x2 – 4 )
b.
*
(x – 2)
(x2– x )
(x + 3)(x – 1 )
(x + 2 )(x – 2)
=
(x – 2)
x(x – 1 )
(x + 3)(x + 2)
=
x

A
C
AC
*
=
B
D
BD
Example A. Simplify
10x
y2
10xy2
2
a.
=
=
*
y3z
5x3
5x3y3z
x2yz
(x2 + 2x – 3 )
(x2 – 4 )
b.
*
(x – 2)
(x2– x )
(x + 3)(x – 1 )
(x + 2 )(x – 2)
=
(x – 2)
x(x – 1 )
(x + 3)(x + 2)
=
x
In the next section, we meet the following type of problems.

Example B. Simplify and expand the answers.
x + 3
a.
(x2– 1)
x – 1

x + 3
a.
(x2– 1)
x – 1
x + 3
=
(x – 1)(x + 1)
(x – 1)

x + 3
a.
(x2– 1)
x – 1
x + 3
=
(x – 1)(x + 1)
(x – 1)
=
(x + 3)(x + 1)

x + 3
a.
(x2– 1)
x – 1
x + 3
=
(x – 1)(x + 1)
(x – 1)
=
(x + 3)(x + 1) = x2 + 4x + 3

x + 3
a.
(x2– 1)
x – 1
x + 3
=
(x – 1)(x + 1)
(x – 1)
=
(x + 3)(x + 1) = x2 + 4x + 3
x – 2
(
x + 1
)
( x – 3)(x + 3)(x + 1)
b.
–
x2– 9
x2– 2x – 3

x + 3
a.
(x2– 1)
x – 1
x + 3
=
(x – 1)(x + 1)
(x – 1)
=
(x + 3)(x + 1) = x2 + 4x + 3
x – 2
(
x + 1
)
( x – 3)(x + 3)(x + 1)
b.
–
x2– 9
x2– 2x – 3
x – 2
=
(x – 3)(x + 3)

x + 3
a.
(x2– 1)
x – 1
x + 3
=
(x – 1)(x + 1)
(x – 1)
=
(x + 3)(x + 1) = x2 + 4x + 3
x – 2
(
x + 1
)
( x – 3)(x + 3)(x + 1)
b.
–
x2– 9
x2– 2x – 3
x – 2
x + 1
=
–
(x – 3)(x + 3)
(x – 3)(x + 1)

x + 3
a.
(x2– 1)
x – 1
x + 3
=
(x – 1)(x + 1)
(x – 1)
=
(x + 3)(x + 1) = x2 + 4x + 3
x – 2
(
x + 1
)
( x – 3)(x + 3)(x + 1)
b.
–
x2– 9
x2– 2x – 3
x – 2
[
x + 1
]
( x – 3)(x + 3)(x + 1)
=
–
(x – 3)(x + 3)
(x – 3)(x + 1)

x + 3
a.
(x2– 1)
x – 1
x + 3
=
(x – 1)(x + 1)
(x – 1)
=
(x + 3)(x + 1) = x2 + 4x + 3
x – 2
(
x + 1
)
( x – 3)(x + 3)(x + 1)
b.
–
x2– 9
x2– 2x – 3
(x + 1)
x – 2
[
x + 1
]
( x – 3)(x + 3)(x + 1)
=
–
(x – 3)(x + 3)
(x – 3)(x + 1)

x + 3
a.
(x2– 1)
x – 1
x + 3
=
(x – 1)(x + 1)
(x – 1)
=
(x + 3)(x + 1) = x2 + 4x + 3
x – 2
(
x + 1
)
( x – 3)(x + 3)(x + 1)
b.
–
x2– 9
x2– 2x – 3
(x + 1)
(x + 3)
x – 2
[
x + 1
]
( x – 3)(x + 3)(x + 1)
=
–
(x – 3)(x + 3)
(x – 3)(x + 1)

x + 3
a.
(x2– 1)
x – 1
x + 3
=
(x – 1)(x + 1)
(x – 1)
=
(x + 3)(x + 1) = x2 + 4x + 3
x – 2
(
x + 1
)
( x – 3)(x + 3)(x + 1)
b.
–
x2– 9
x2– 2x – 3
(x + 1)
(x + 3)
x – 2
[
x + 1
]
( x – 3)(x + 3)(x + 1)
=
–
(x – 3)(x + 3)
(x – 3)(x + 1)
=
(x – 2)(x + 1) – (x + 1)(x + 3)

x + 3
a.
(x2– 1)
x – 1
x + 3
=
(x – 1)(x + 1)
(x – 1)
=
(x + 3)(x + 1) = x2 + 4x + 3
x – 2
(
x + 1
)
( x – 3)(x + 3)(x + 1)
b.
–
x2– 9
x2– 2x – 3
(x + 1)
(x + 3)
x – 2
[
x + 1
]
( x – 3)(x + 3)(x + 1)
=
–
(x – 3)(x + 3)
(x – 3)(x + 1)
=
(x – 2)(x + 1) – (x + 1)(x + 3)
= (x – 2)(x + 1) + (–x –1)(x + 3)

x + 3
a.
(x2– 1)
x – 1
x + 3
=
(x – 1)(x + 1)
(x – 1)
=
(x + 3)(x + 1) = x2 + 4x + 3
x – 2
(
x + 1
)
( x – 3)(x + 3)(x + 1)
b.
–
x2– 9
x2– 2x – 3
(x + 1)
(x + 3)
x – 2
[
x + 1
]
( x – 3)(x + 3)(x + 1)
=
–
(x – 3)(x + 3)
(x – 3)(x + 1)
=
(x – 2)(x + 1) – (x + 1)(x + 3)
= (x – 2)(x + 1) + (–x –1)(x + 3)
= x2– x – 2 – x2 – 4x – 3

x + 3
a.
(x2– 1)
x – 1
x + 3
=
(x – 1)(x + 1)
(x – 1)
=
(x + 3)(x + 1) = x2 + 4x + 3
x – 2
(
x + 1
)
( x – 3)(x + 3)(x + 1)
b.
–
x2– 9
x2– 2x – 3
(x + 1)
(x + 3)
x – 2
[
x + 1
]
( x – 3)(x + 3)(x + 1)
=
–
(x – 3)(x + 3)
(x – 3)(x + 1)
=
(x – 2)(x + 1) – (x + 1)(x + 3)
= (x – 2)(x + 1) + (–x –1)(x + 3)
= x2– x – 2 – x2 – 4x – 3
= –5x – 5

Division Rule for Rational Expressions
C
A
÷
B
D

C
A
AD
÷
=
B
D
BC
Reciprocate

C
A
AD
÷
=
B
D
BC
Reciprocate
We convert division by an expression to multiplying by its reciprocal.

C
A
AD
÷
=
B
D
BC
Reciprocate
We convert the act of dividing by an quantity xto multiplying by its reciprocal:1/x. Then we factor and reduce the product.

C
A
AD
÷
=
B
D
BC
Reciprocate
Example C. Simplify
(2x – 6)
(9 – x2)
÷
(x + 3)
(x2+ 2x – 3)

C
A
AD
÷
=
B
D
BC
Reciprocate
Example C. Simplify
(2x – 6)
(9 – x2)
÷
(x + 3)
(x2+ 2x – 3)
(2x – 6)
(x2+ 2x – 3)
=
*
(x + 3)
(9 – x2)

C
A
AD
÷
=
B
D
BC
Reciprocate
Example C. Simplify
(2x – 6)
(9 – x2)
÷
(x + 3)
(x2+ 2x – 3)
(2x – 6)
(x2+ 2x – 3)
=
*
(x + 3)
(9 – x2)
2(x – 3)
=
(x + 3)

C
A
AD
÷
=
B
D
BC
Reciprocate
Example C. Simplify
(2x – 6)
(9 – x2)
÷
(x + 3)
(x2+ 2x – 3)
(2x – 6)
(x2+ 2x – 3)
=
*
(x + 3)
(9 – x2)
2(x – 3)
(x + 3)(x – 1)
=
(x + 3)
(3 – x)(3 + x)

C
A
AD
÷
=
B
D
BC
Reciprocate
Example C. Simplify
(2x – 6)
(9 – x2)
(9 – x2)
÷
(x + 3)
(x2+ 2x – 3)
(2x – 6)
(x2+ 2x – 3)
=
*
(x + 3)
(9 – x2)
2(x – 3)
(x + 3)(x – 1)
=
*
(x + 3)
(3 – x)(3 + x)

C
A
AD
÷
=
B
D
BC
Reciprocate
Example C. Simplify
(2x – 6)
(9 – x2)
÷
(x + 3)
(x2+ 2x – 3)
(2x – 6)
(x2+ 2x – 3)
=
*
(x + 3)
(9 – x2)
(–1)
2(x – 3)
(x + 3)(x – 1)
=
*
(x + 3)
(3 – x)(3 + x)

C
A
AD
÷
=
B
D
BC
Reciprocate
Example C. Simplify
(2x – 6)
(9 – x2)
÷
(x + 3)
(x2+ 2x – 3)
(2x – 6)
(x2+ 2x – 3)
=
*
(x + 3)
(9 – x2)
(–1)
2(x – 3)
(x + 3)(x – 1)
=
*
(x + 3)
(3 – x)(3 + x)
–2(x – 1)
=
(3 + x)

Besides the expanded form and factored forms rational expressions may be split into sums or differences.

There are two common ways to do this.

I. Split off the numerator term by term.

Example D. Break up the numerators as the sums or differences and simplify each term.
(2x – 6)
a.
=
(x + 3)
(2x – 6)
b.
=
3x2

(2x – 6)
2x
6
a.
=
–
(x + 3)
(x + 3)
(x + 3)
(2x – 6)
b.
=
3x2

(2x – 6)
2x
6
a.
=
–
(x + 3)
(x + 3)
(x + 3)
(2x – 6)
2x
6
b.
=
–
3x2
3x2
3x2

(2x – 6)
2x
6
a.
=
–
(x + 3)
(x + 3)
(x + 3)
(2x – 6)
2x
2
6
2
b.
=
–
–
=
3x2
3x2
3x2
x2
3x

(2x – 6)
2x
6
a.
=
–
(x + 3)
(x + 3)
(x + 3)
(2x – 6)
2x
2
6
2
b.
=
–
–
=
3x2
3x2
3x2
x2
3x
II. Long Division
Long division is the extension of the long division of numbers from grade schools and it is for the division of polynomials in one variable.

(2x – 6)
2x
6
a.
=
–
(x + 3)
(x + 3)
(x + 3)
(2x – 6)
2x
2
6
2
b.
=
–
–
=
3x2
3x2
3x2
x2
3x
II. Long Division
Long division is the extension of the long division of numbers from grade schools and it is for the division of polynomials in one variable. Specifically, long division gives relevant results only when the degree of the numerator is the same or more than the degree of the denominator.

Let’s look at the example 125/8 or 125 ÷ 8 by long division.

i. Put the problem in the long division format with the “bottom-out” and move from left to right until there is enough to enter a quotient.

)
8
125

1
)
8
125

1
)
8
125
ii. Multiply the quotient back into the problem and subtract the results from the numerator. Bring down the remaining terms from the numerator.
8
45

1
)
8
125
8
45
iii. Repeat steps i and ii until no more quotient may be entered.

15
)
8
125
8
45
40
5

15
)
8
125
8
45
40
5
iii. Repeat steps i and ii until no more quotient may be entered. Then we may put the fraction N/D in the following mixed form:
R
N
where Q is the quotient
= Q +
D
D
and that R (the remainder) is smaller then D (no more quotient).

15
)
8
125
8
45
40
5
5
125
= 15 +
8
8
R
N
= Q +
D
D
and that R (the remainder) is smaller then D (no more quotient).

(2x – 6)
Example E. Divide using long division
(x + 3)

(2x – 6)
(x + 3)
i. Put the problem in the long division format with the “bottom-out” and
enter the quotients of the leading terms.

(2x – 6)
(x + 3)
)
x + 3
2x – 6
Make sure the terms
are in order.

(2x – 6)
(x + 3)
)
x + 3
2x – 6
enter the quotients of the leading terms 2x/x = 2

(2x – 6)
(x + 3)
2
)
x + 3
2x – 6
enter the quotients of the leading terms 2x/x = 2

(2x – 6)
(x + 3)
2
)
x + 3
2x – 6

(2x – 6)
(x + 3)
2
)
x + 3
2x – 6
2x + 6

(2x – 6)
(x + 3)
2
)
x + 3
2x – 6
–)
2x + 6
–12

(2x – 6)
(x + 3)
2
)
x + 3
2x – 6
–)
2x + 6
–12
Stop. No more quotient since
x can’t going into 12.

(2x – 6)
(x + 3)
2
)
x + 3
2x – 6
–)
2x + 6
–12
R
N
and the remainder R
= Q +
D
D
has smaller degree then denominator D (no more quotient).

(2x – 6)
(x + 3)
2
)
x + 3
2x – 6
–)
2x + 6
–12
Hence we may write
(2x – 6)
(x + 3)
R
N
and the remainder R
= Q +
D
D

(2x – 6)
(x + 3)
Q
2
)
x + 3
2x – 6
R
–)
2x + 6
–12
Hence we may write
(2x – 6)
(x + 3)
R
N
and the remainder R
= Q +
D
D

(2x – 6)
(x + 3)
Q
2
)
x + 3
2x – 6
R
–)
2x + 6
–12
Hence we may write
(2x – 6)
(x + 3)
Q
R
12
= 2 –
x + 3
R
N
and the remainder R
= Q +
D
D

x2 – 6x + 3
Example F. Divide using long division
x – 2
R
N
and the remainder R
= Q +
D
D

x2 – 6x + 3
x – 2
)
x + 3
x2 – 6x + 3
Make sure the terms
are in order.
R
N
and the remainder R
= Q +
D
D

x2 – 6x + 3
x – 2
)
x + 3
x2 – 6x + 3
R
N
and the remainder R
= Q +
D
D

x2 – 6x + 3
x – 2
x
)
x + 3
x2 – 6x + 3
R
N
and the remainder R
= Q +
D
D

x2 – 6x + 3
x – 2
x
)
x + 3
x2 – 6x + 3
x2 + 3x
R
N
and the remainder R
= Q +
D
D

x2 – 6x + 3
x – 2
x
)
x + 3
x2 – 6x + 3
–)
x2 + 3x
–9x + 3
R
N
and the remainder R
= Q +
D
D

x2 – 6x + 3
x – 2
x – 9
)
x + 3
x2 – 6x + 3
–)
x2 + 3x
–9x + 3
R
N
and the remainder R
= Q +
D
D

x2 – 6x + 3
x – 2
x – 9
)
x + 3
x2 – 6x + 3
–)
x2 + 3x
–9x + 3
–9x – 27
R
N
and the remainder R
= Q +
D
D

x2 – 6x + 3
x – 2
x – 9
)
x + 3
x2 – 6x + 3
–)
x2 + 3x
–9x + 3
–)
–9x – 27
30
R
N
and the remainder R
= Q +
D
D

x2 – 6x + 3
x – 2
x – 9
)
x + 3
x2 – 6x + 3
–)
x2 + 3x
–9x + 3
–)
–9x – 27
30
Stop. No more quotient since x can’t going into 30. Hence 30 is the remainder.
R
N
and the remainder R
= Q +
D
D

x – 2
x2 – 6x + 3
x – 2
x – 9
)
x + 3
x2 – 6x + 3
–)
x2 + 3x
–9x + 3
–)
–9x – 27
30
Hence
x2 – 6x + 3
x – 2
=
R
N
and the remainder R
= Q +
D
D

x – 2
x2 – 6x + 3
x – 2
x – 9
)
x + 3
x2 – 6x + 3
–)
x2 + 3x
–9x + 3
–)
–9x – 27
30
Hence
x2 – 6x + 3
x – 2
30
= x – 9 +
x + 3
R
N
and the remainder R
= Q +
D
D

Ex A. Simplify. Do not expand the results.
3
3
5
2
3.
12x6
5x3
4.
1.
10x
5x3
2.
*
*
10x
*
10x
*
6x14
5x3
56x6
63
75x
42
15x
16
10
18
7.
8.
5.
6.
*
*
*
*
8x5
25x3
25x4
5x3
27
49
4
9x4
6 – 4x
x – 2
2x – 4
5x + 10
10.
*
9.
3x – 2
2x + 4
*
2x + 4
3x – 6
4 – x
x + 4
9x – 12
2 – x
12.
*
11.
*
–x – 4
x – 4
2x – 4
8 – 6x
42 – 6x
4 – 2x
3x – 9
3 – x
14.
*
13.
*
–2x + 14
–7x + 14
15x – 5
5 – 15x
(x2 + 2x – 3 )
(x2 – 2x – 3)
(x2+ x – 2 )
(x2 – 4 )
16.
15.
*
(x2–9)
(x2–x – 2 )
*
(x – 2)
(x2– x)
(x2 + 5x – 6 )
(x2 – 5x + 6)
(x2– x – 2 )
(x2+ x )
18.
17.
*
(x2+ 5x + 6)
(x2–5x – 6 )
*
(x2–1)
(x2+ 2x + 1)
(– x2 + 6 – x )
(6 – x2 – x)
(x2– 3x – 4 )
(x2– 3x + 2)
20.
19.
*
(x2+ 5x + 6)
(x2– x – 12 )
*
(x2–1)
(x2– 2x – 8)

Ex. A. Simplify. Do not expand the results.
(3x2 – 2x – 1)
(x2+ x – 2 )
(2x2 + x – 1 )
(4x2– 1)
22.
21.
÷
÷
(1 – 9x2)
(x2+ 4x + 4)
(1 –2x)
(2x2– x )
(x + 1 – 6x2)
(2x2+ x – 1 )
(x2–x + 2)
(3x2+ 4x + 1)
24.
23.
÷
÷
(3x2 – x – 2)
(–x2– 4)
(x2– 5x – 6)
(–x – 3x2)
(–x3 + 9x )
(x2+ 6x + 9)
(x3 – 4x)
(x2+ 2)
25.
26.
÷
÷
(–x2 + 4x – 4)
(–x + 2)
(x2+ 3x)
(–3x2 – 9x)
Ex. B. Multiply, expand and simplify the results.
x + 3
x – 3
2x + 3
27.
28.
29.
(x2– 1)
(x2–4)
(x2– 1)
x +1
x –2
1 – x
3 – 2x
3 – 2x
30.
31.
(x + 2)(x +1)
(3x + 2)(1 – 2x)
x + 2
2x – 1
x – 2
(
x + 1
)
32.
+
( x – 3)(x + 3)
x –3
x + 3
2x – 1
(
x + 2
)
( 2x – 3)(x + 2)
33.
–
x + 2
2x – 3

Ex B. Multiply, expand and simplify the results.
4 – x
(
x – 1
)
34.
–
( x – 3)(2x + 3)
x –3
2x + 3
3 – x
(
2x + 3
)
(x – 3)(x + 2)
35.
–
x + 2
x – 3
3 – 4x
(
1 – 2x
)
36.
–
( x + 3)(x + 1)
x + 1
x + 3
5x – 7
(
4 – 5x
)
(x – 3)(x + 5)
37.
–
x + 5
x – 3
x – 2
(
x + 1
)
( x – 3)(x + 3)(x + 1)
38.
–
x2– 9
x2– 2x – 3
x + 3
(
2x + 1
)
( x – 2)(x + 2)(x – 1)
39.
–
x2–4
x2+ x – 2
x – 1
(
x + 1
)
( x – 3)(x + 2)(x + 1)
40.
–
x2– x – 6
x2– 2x – 3
x + 2
(
2x + 1
)
( x – 3)(x + 3)(x – 1)
41.
–
x2– 4x +3
x2+2x – 3

Ex. C. Break up the following expressions as sums and differences of fractions.
x2+ 4x – 6
x2 – 4
x2– 4
43.
44.
2x2
42.
x2
2x
12x3 – 9x2 + 6x
x8– x6– x4
x8– x6– x4
46.
47.
45.
x
3x
x2
Ex D. Use long division and write each rational expression in
the form of Q + .
R
D
2x + 6
3x – 5
4x + 3
48.
49.
50.
x + 2
x – 2
x – 1
5x – 4
3x + 8
–4x – 5
51.
52.
53.
x – 3
2 – x
1 – x
(x2+ x – 2 )
(3x2 – 3x – 2 )
(2x2 + x – 3 )
54.
55.
56.
(x –1)
(x + 2)
(x – 2)
(–x2 + 4x – 3 )
(5x2 – 1 )
(4x2 + 2 )
57.
58.
59.
(x –3)
(x –4)
(x + 3)

1 4 multiplication and division of rational expressions

by math123b on Aug 28, 2011

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1 4 multiplication and division of rational expressions — Presentation Transcript