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## Multiplying and Dividing Radical Expressions - Sample Math Practice Problems

The math problems below can be generated by MathScore.com, a math practice program for schools and individual families. References to complexity and mode refer to the overall difficulty of the problems as they appear in the main program. In the main program, all problems are automatically graded and the difficulty adapts dynamically based on performance. Answers to these sample questions appear at the bottom of the page. This page does not grade your responses.

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### Complexity=1, Mode=mul

Solve and simplify.

1.
 √5 • √45 =
2.
 √6 • √30 =

### Complexity=2, Mode=mul

Solve and simplify.

1.
 70p 8
50p =
2.
45j
 27j 4
=

### Complexity=1, Mode=div

Solve and simplify. Leave fractions as improper fractions.

1.
=
2.
=

### Complexity=2, Mode=div

Solve and simplify. Leave fractions as improper fractions.

1.
=
2.
=

### Complexity=3, Mode=mul

Solve and simplify.

1.
 8b 5
 6a 7b 8
=
2.
 4tu 7
 5t 2u 5
=

### Complexity=3, Mode=div

Solve and simplify. Leave fractions as improper fractions.

1.
=
2.

=

### Complexity=1, Mode=mul

Solve and simplify.

1
 √5 • √45 =
Solution
 √5 • √45 = √5 • 45 = √5 • 5 • 9 = 5 • √3 • 3 = 5 • 3 = 15
2
 √6 • √30 =
Solution
 √6 • √30 = √6 • 30 = √6 • 6 • 5 = 6√5

### Complexity=2, Mode=mul

Solve and simplify.

1
 70p 8
50p =
Solution
 70p 8
 50p
=
 10 • 7 • p 8
 10 • 5 • p
=
 5 • 7 • 10 • 10
 p 8 • p
=
10√5 • 7
 p 9
=
 10√35 • p 4√ p
=10p 435p
2
45j
 27j 4
=
Solution
 45j
 27j 4
=
 9 • 5 • j
 9 • 3 • j 4
=
 3 • 5 • 9 • 9
 j • j 4
=
9√3 • 5
 j 5
=
 9√15 • j 2√ j
=9j 215j

### Complexity=1, Mode=div

Solve and simplify. Leave fractions as improper fractions.

1
=
Solution
 = = =
2
=
Solution
 = =

### Complexity=2, Mode=div

Solve and simplify. Leave fractions as improper fractions.

1
=
Solution
The numerator and denominator are both divisible by 2
=

=

=
=
 •
=
2
=
Solution
=
 •
=

### Complexity=3, Mode=mul

Solve and simplify.

1
 8b 5
 6a 7b 8
=
Solution
 8b 5
 6a 7b 8
=
 2 • 4 • b 5
 2 • 3 • a 7 • b 8
=
 2 • 2 • 3 • 4
 b 5 • b 8
 a 7
=
2√3 • 4
 b 13
a 3 a
=
 (2 • 2)√3 • b 6√ b • a 3√ a
=
 4√3 • b 6√ b • a 3√ a
=4a 3b 63ab
2
 4tu 7
 5t 2u 5
=
Solution
 4tu 7
 5t 2u 5
=
 4 • t • u 7
 5 • t 2 • u 5
=
 4 • 5
 t • t 2
 u 7 • u 5
=
2√5
 t 3
 u 12
=
 2√5 • t√ t • u 6
=2tu 65t

### Complexity=3, Mode=div

Solve and simplify. Leave fractions as improper fractions.

1
=
Solution
=
=
 •
=
2

=
Solution
The numerator and denominator are both divisible by q 2

=

=   