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Simplifying the expression six a plus the product of five and the binomial a plus three, using the distributive property, and combining like terms. See the longdesc for a full description.

4 x + 9 ( x 2 6 x 2 ) + 5 Remove parentheses . 4 x + 9 x 2 54 x 18 + 5 Combine like terms . 50 x + 9 x 2 13

By convention, the terms in an expression are placed in descending order with the highest degree term appearing first. Numerical terms are placed at the right end of the expression. The commutative property of addition allows us to change the order of the terms.

9 x 2 50 x 13

2 + 2 [ 5 + 4 ( 1 + a ) ]
Eliminate the innermost set of parentheses first.

2 + 2 [ 5 + 4 + 4 a ]

By the order of operations, simplify inside the parentheses before multiplying (by the 2).

2 + 2 [ 9 + 4 a ] Remove this set of parentheses . 2 + 18 + 8 a Combine like terms . 20 + 8 a Write in descending order . 8 a + 20

x ( x 3 ) + 6 x ( 2 x + 3 )
Use the rule for multiplying powers with the same base.

x 2 3 x + 12 x 2 + 18 x Combine like terms . 13 x 2 + 15 x

Practice set c

Simplify each of the following expressions by using the distributive property and combining like terms.

4 ( x + 6 ) + 3 ( 2 + x + 3 x 2 ) 2 x 2

7 x 2 + 7 x + 30

7 ( x + x 3 ) 4 x 3 x + 1 + 4 ( x 2 2 x 3 + 7 )

5 x 3 + 4 x 2 + 6 x + 29

5 ( a + 2 ) + 6 a 7 + ( 8 + 4 ) ( a + 3 a + 2 )

59 a + 27

x ( x + 3 ) + 4 x 2 + 2 x

5 x 2 + 5 x

a 3 ( a 2 + a + 5 ) + a ( a 4 + 3 a 2 + 4 ) + 1

2 a 5 + a 4 + 8 a 3 + 4 a + 1

2 [ 8 3 ( x 3 ) ]

6 x + 34

x 2 + 3 x + 7 [ x + 4 x 2 + 3 ( x + x 2 ) ]

50 x 2 + 31 x

Exercises

For the following problems, simplify each of the algebraic expressions.

x + 3 x

4 x

4 x + 7 x

9 a + 12 a

21 a

5 m 3 m

10 x 7 x

3 x

7 y 9 y

6 k 11 k

5 k

3 a + 5 a + 2 a

9 y + 10 y + 2 y

21 y

5 m 7 m 2 m

h 3 h 5 h

7 h

a + 8 a + 3 a

7 a b + 4 a b

11 a b

8 a x + 2 a x + 6 a x

3 a 2 + 6 a 2 + 2 a 2

11 a 2

14 a 2 b + 4 a 2 b + 19 a 2 b

10 y 15 y

5 y

7 a b 9 a b + 4 a b

210 a b 4 + 412 a b 4 + 100 a 4 b ( Look closely at the exponents. )

622 a b 4 + 100 a 4 b

5 x 2 y 0 + 3 x 2 y + 2 x 2 y + 1 , y 0 ( Look closely at the exponents. )

8 w 2 12 w 2 3 w 2

7 w 2

6 x y 3 x y + 7 x y 18 x y

7 x 3 2 x 2 10 x + 1 5 x 2 3 x 3 12 + x

4 x 3 7 x 2 9 x 11

21 y 15 x + 40 x y 6 11 y + 7 12 x x y

1 x + 1 y 1 x 1 y + x y

x y

5 x 2 3 x 7 + 2 x 2 x

2 z 3 + 15 z + 4 z 3 + z 2 6 z 2 + z

2 z 3 5 z 2 + 16 z

18 x 2 y 14 x 2 y 20 x 2 y

9 w 5 9 w 4 9 w 5 + 10 w 4

18 w 5 + w 4

2 x 4 + 4 x 3 8 x 2 + 12 x 1 7 x 3 1 x 4 6 x + 2

17 d 3 r + 3 d 3 r 5 d 3 r + 6 d 2 r + d 3 r 30 d 2 r + 3 7 + 2

16 d 3 r 24 d 2 r 2

a 0 + 2 a 0 4 a 0 , a 0

4 x 0 + 3 x 0 5 x 0 + 7 x 0 x 0 , x 0

8

2 a 3 b 2 c + 3 a 2 b 2 c 0 + 4 a 2 b 2 a 3 b 2 c , c 0

3 z 6 z + 8 z

5 z

3 z 2 z + 3 z 3

6 x 3 + 12 x + 5

6 x 3 + 12 x + 5

3 ( x + 5 ) + 2 x

7 ( a + 2 ) + 4

7 a + 18

y + 5 ( y + 6 )

2 b + 6 ( 3 5 b )

28 b + 18

5 a 7 c + 3 ( a c )

8 x 3 x + 4 ( 2 x + 5 ) + 3 ( 6 x 4 )

31 x + 8

2 z + 4 a b + 5 z a b + 12 ( 1 a b z )

( a + 5 ) 4 + 6 a 20

10 a

( 4 a + 5 b 2 ) 3 + 3 ( 4 a + 5 b 2 )

( 10 x + 3 y 2 ) 4 + 4 ( 10 x + 3 y 2 )

80 x + 24 y 2

2 ( x 6 ) + 5

1 ( 3 x + 15 ) + 2 x 12

5 x + 3

1 ( 2 + 9 a + 4 a 2 ) + a 2 11 a

1 ( 2 x 6 b + 6 a 2 b + 8 b 2 ) + 1 ( 5 x + 2 b 3 a 2 b )

3 a 2 b + 8 b 2 4 b + 7 x

After observing the following problems, can you make a conjecture about 1 ( a + b ) ?
1 ( a + b ) =

Using the result of problem 52, is it correct to write
( a + b ) = a + b ?

yes

3 ( 2 a + 2 a 2 ) + 8 ( 3 a + 3 a 2 )

x ( x + 2 ) + 2 ( x 2 + 3 x 4 )

3 x 2 + 8 x 8

A ( A + 7 ) + 4 ( A 2 + 3 a + 1 )

b ( 2 b 3 + 5 b 2 + b + 6 ) 6 b 2 4 b + 2

2 b 4 + 5 b 3 5 b 2 + 2 b + 2

4 a a ( a + 5 )

x 3 x ( x 2 7 x 1 )

3 x 3 + 21 x 2 + 4 x

a b ( a 5 ) 4 a 2 b + 2 a b 2

x y ( 3 x y + 2 x 5 y ) 2 x 2 y 2 5 x 2 y + 4 x y 2

x 2 y 2 3 x 2 y x y 2

3 h [ 2 h + 5 ( h + 2 ) ]

2 k [ 5 k + 3 ( 1 + 7 k ) ]

52 k 2 + 6 k

8 a [ 2 a 4 a b + 9 ( a 5 a b ) ]

6 { m + 5 n [ n + 3 ( n 1 ) ] + 2 n 2 } 4 n 2 9 m

128 n 2 90 n 3 m

5 [ 4 ( r 2 s ) 3 r 5 s ] + 12 s

8 { 9 [ b 2 a + 6 c ( c + 4 ) 4 c 2 ] + 4 a + b } 3 b

144 c 2 112 a + 77 b + 1728 c

5 [ 4 ( 6 x 3 ) + x ] 2 x 25 x + 4

3 x y 2 ( 4 x y + 5 y ) + 2 x y 3 + 6 x 2 y 3 + 4 y 3 12 x y 3

18 x 2 y 3 + 5 x y 3 + 4 y 3

9 a 3 b 7 ( a 3 b 5 2 a 2 b 2 + 6 ) 2 a ( a 2 b 7 5 a 5 b 12 + 3 a 4 b 9 ) a 3 b 7

8 ( 3 a + 2 )

24 a 16

4 ( 2 x 3 y )

4 x y 2 [ 7 x y 6 ( 5 x y 2 ) + 3 ( x y + 1 ) + 1 ]

24 x 2 y 4 16 x 2 y 3 + 104 x y 2

Exercises for review

( [link] ) Simplify ( x 10 y 8 z 2 x 2 y 6 ) 3 .

( [link] ) Find the value of 3 ( 4 9 ) 6 ( 3 ) 1 2 3 .

4

( [link] ) Write the expression 42 x 2 y 5 z 3 21 x 4 y 7 so that no denominator appears.

( [link] ) How many ( 2 a + 5 ) ' s are there in 3 x ( 2 a + 5 ) ?

3 x

( [link] ) Simplify 3 ( 5 n + 6 m 2 ) 2 ( 3 n + 4 m 2 ) .

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Source:  OpenStax, Algebra i for the community college. OpenStax CNX. Dec 19, 2014 Download for free at http://legacy.cnx.org/content/col11598/1.3
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