Example Add. Simplify the result, if possible.

a) b)

Solutiona)

b)

2 23 4 9 3

3 1 3 1

x x x x

x x

2 2

9 3

36 36

x

x x

2 2 2 23 4 9 3 (3 4 9) ( 3

3 1 3 1 1

)

3x x x

x x x x x x x x

24 5 12

3 1

x x

x

Combining like terms

2 2 2

9 3 6

36 36 36

x x

x x x

6

( 6)( 6)

x

x x

1 ( 6)x

( 6) ( 6)x x 1

6x

Factoring

Combining like terms in the numerator

Example

Subtract and, if possible, simplify:a) b)

Solution

a)

5 4

3 3

x x

x x

2 30

6 6

x x

x x

(5 5

3 3

)4 4

3

x x x x

x x x

5

3

4x

x

x

4 4

3

x

x

The parentheses are needed to make sure that we practice safe math.

Removing the parentheses and changing the signs (using the distributive law)

Combining like terms

Example continued

b) 2 230 30(

6

)

6 6

x x x x

x x x

2

6

30x

x

x

( 5)

6

6)(x x

x

( 6)x

( 5)

6

x

x

5x

Removing the parentheses (using the distributive law)

Factoring, in hopes of simplifying

Removing the clever form of 1

Example

For each pair of polynomials, find the least common multiple.a) 16a and 24bb) 24x4y4 and 6x6y2

c) x2 4 and x2 2x 8Solutiona) 16a = 2 2 2 2 a 24b = 2 2 2 3 b

The LCM = 2 2 2 2 a 3 b

The LCM is 24 3 a b, or 48ab

16a is a factor of the LCM

24b is a factor of the LCM

Example continued

b) 24x4y4 = 2 2 2 3 x x x x y y y y 6x6y2 = 2 3 x x x x x x y y

LCM = 2 2 2 3 x x x x y y y y x x

Note that we used the highest power of each factor. The LCM is 24x6y4

c) x2 4 = (x 2)(x + 2) x2 2x 8 = (x + 2)(x 4)

LCM = (x 2)(x + 2)(x 4)

x2 4 is a factor of the LCM

x2 2x 8 is a factor of the LCM

Example

For each group of polynomials, find the least common multiple.a) 15x, 30y, 25xyz b) x2 + 3, x + 2, 7

Solution

a) 15x = 3 5 x 30y = 2 3 5 y 25xyz = 5 5 x y z LCM = 2 3 5 5 x y z The LCM is 2 3 52 x y z or 150xyz

b) Since x2 + 3, x + 2, and 7 are not factorable, the LCM is their product: 7(x2 + 3)(x + 2).

Solution1. First, we find the LCD:

9 = 3 312 = 2 2 3

2. Multiply each expression by the appropriate number to get the LCD.

Example Add: 24 5

.9 12

x x

2 24 5 4 5

9 12 3 3 2 2 3

x x x x

LCD = 2 2 3 3 = 36

2 4

4

4 5

3 3 2 2 3

3

3

x x

216 15

36 36

x x

=

Solution First, we find the LCD:

a2 4 = (a 2)(a + 2) a2 2a = a(a 2)

We multiply by a form of 1 to get the LCD in each expression:

Example Add: 2 2

3 2.

4 2

a

a a a

LCD = a(a 2)(a + 2).

2 2

3 2 3 2

4 2 ( 2)( 2) ( 2 2)

2a a

a a a a a a

a

a

a

aa

23 2 4

( 2)( 2) ( 2)( 2)

a a

a a a a a a

23 2 4

( 2)( 2)

a a

a a a

3a2 + 2a + 4 will not factor so we are done.

Solution First, we find the LCD. It is just the product of the denominators: LCD = (x + 4)(x + 6).

We multiply by a form of 1 to get the LCD in each expression. Then we subtract and try to simplify.

Example Subtract: 2 1

.4 6

x x

x x

2 1 2 1

4 6 4 4

6 4

66

x x x x

x x x x x

x x

x

2 28 12 3 4

( 4)( 6) ( 4)( 6)

x x x x

x x x x

Multiplying out numerators

2 28 12 3 4

( 4)( 6) ( 4)( 6)

x x x x

x x x x

2 28 12 3 4( )

( 4)( 6)

x x x x

x x

When subtracting a numerator with more than one term, parentheses are important, practice safe math.

2 2 38 12

( ( 6)

4

4)

x x

x

xx

x

5 16

( 4)( 6)

x

x x

Solution

Example Add:

2 2

1 4.

4 4 3 10

x

x x x x

2 2

1 4 1 4

4 4 3 10 ( 2)( 2) ( 2)( 5)

x x

x x x x x x x x

5

5

21 4

( 2)( 2) ( 2)( 5) 2

x

x

x

x

x

x x x x

2 6 5 4 8

( 2)( 2)( 5) ( 2)( 2)( 5)

x x x

x x x x x x

2 6 5 4 8

( 2)( 2)( 5)

x x x

x x x

Adding numerators

2 10 3

( 2)( 2)( 5)

x x

x x x

Continued