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
Top Related