The Mathematics 11
Competency Test
Multiplication and Division With
Square Roots
We have already looked in some detail at multiplication and division with numerical square roots.
The rules for multiplication and division with algebraic square roots are exactly the same as those
for numerical square roots – in fact, we stated those rules there in an algebraic form:
a
b 
a b
(property 1)
and
a 
a
b 
a
b

b
(property 2)
where ‘a’ and ‘b’ stand for any valid mathematical expression.
When expressions involving square roots are to be multiplied or divided, just use the above rules
to combine square roots as necessary. Then simplify the resulting expression as much as
possible using methods of simplification already described and illustrated extensively in the
preceding documents in this series.
The large collection of examples following below illustrate this general strategy, but also provide
some examples for your own practice.
Example 1: Simplify
10 x 3
5x5 .
solution:
Using property (1) above,
10x3
5x5 
10x 5x 
3
5

50x 35

50x 8

52 2
 5x 4
x 
4 2

52
x 
4 2
2
2
as the final answer.
David W. Sabo (2003)
Multiplication and Division With Square Roots
Page 1 of 10
8x3 y
Example 2: Simplify
32 xy 3 .
solution:
Using property (1) above, we get
8 32 x 31y 13
32xy 3 
8x 3 y

256x 4 y 4

16   x 2   y 2 

162
2
2
x 
2
2
2
y 
2
2
 16x 2 y 2
Example 3: Simplify

.
2
3 2xy
solution:
You could regard this problem as a product
3
  3
2
2xy
2xy
3
2xy

and use the procedures illustrated in the previous examples. However, an even faster way to get
the answer is to distribute the power 2 over all factors in the brackets:
3
2xy

2
 3
2

2xy

2
  9  2xy 
 18xy
as the final answer. Here we have used the basic fact that
David W. Sabo (2003)

a

2
 a.
Multiplication and Division With Square Roots
Page 2 of 10
Example 4: Simplify
5x
 2y
10y
15 x

solution:
5x
10y
 2y
15 x
  5x  
 10xy
10y
  2y  
15 x

10y 15x 
 10xy 2 5 y 3 5 x
 10 xy 52 6 xy
 10 xy 52
6 xy
 10 xy  5  6 xy
 50xy 6xy
as the final simplified answer.
x
Example 5: Simplify
2x 3
5x5 .
solution:
This is a product of three square root factors, and so does not fit property (1) at the beginning of
this document precisely. However, we can apply property (1) in a stepwise fashion
x
2x 3 
x
2x 3
 x   2x 3 

2x 4
So
5x 5 



2x 3
x

2x 4


5x 5
5x5

which now does match property (1). Continuing

2x 4

David W. Sabo (2003)
5x 5

 2x  5 x 
4
5
Multiplication and Division With Square Roots
Page 3 of 10

10x 4 5

10x 9

10 x 4

x 
 
4
2
2
x
10 x
 x 4 10x
as the final simplified result.
The obvious implication of this stepwise application of property (1) is that property (1) can be
extended to products of any number of square root factors.
6 x 2  24 x
Example 6: Simplify
18 x 4  72 x 3 .
solution:
We can simply go ahead and apply property (1), getting
6x 2  24x
 6x
18x 4  72x 3 
2

 24 x 18 x 4  72x 3

But now, we need to recall that simplification of square roots involves finding perfect square
factors in the square root, so there is no point in expanding the product of the two binomials in
this square root. Instead, we check for the possibility of further factoring.
6x2 + 24x = 6x(x + 4)
and
18x4 + 72x3 = 18x3(x + 4)
Thus
 6x
2

 24x 18x 4  72x 3

 6x  x  4 18x 3   x  4 

3   2 32  x13  x  4 

2

 2  3  x   x  4   3 
2
2
2
2
   x  4
  2  3  x 2
David W. Sabo (2003)
2
2
3
Multiplication and Division With Square Roots
Page 4 of 10
 6x 2  x  4 3
Example 7: Expand and simplify:
1
x

2
.
solution:
At the start, this example involves the square of a binomial. You can use either of the two
approaches presented earlier for multiplying one binomial by another binomial. The result here is
1
  1
2
x
 1
x
1
x

x 
x 

x

2
 1 2 x  x
as the final answer.
Two errors are commonly made with this type of problem on the BCIT Mathematics 11
Competency Test.
(1) People sometimes just square each term of the binomial:
1
x

2
 12 

x

2
 1 x
But, you know that this is not the correct way to expand what amounts to the product of two
binomials.
(2) People sometimes forget the obvious simplification that
1
x

x

2
 12  2 x 

x

2
 12  2 x 

x

2

 1 2 x 
x


x
2

 x , leaving their answer as
2
instead of
1
David W. Sabo (2003)
2
 1 2 x  x
Multiplication and Division With Square Roots
Page 5 of 10
2  3
Example 8: Expand and simplify
y
5  4
y
.
solution:
This example is very similar to the problem in Example 7 above – the product of two binomials,
each of which contains a square root. We get
2  3
y
5  4
y


  2 5    2 4 y
  3
y
 5  3

y

 10  8 y  15 y  12
y
 4
y

2
 10  7 y  12y
as the final result.
2
Example 9: Expand and simplify:
x 3 y
 2
x 3 y
.
solution:
2
 2
x 3 y

 2 x
4

x
x 3 y
 2

2
x
  2
6 x

x
3
y 6 x
y
   3
y
2

y

y 9
x
   3
y
3
y

2
 4 x  9y
as the final result. The two middle terms in the second last line are identical but of opposite sign,
and so cancel each other. This sort of pattern of binomial factors, where each binomial contains
square roots but their product contains no square roots, is the basis for methods to rationalize
denominators of fractions which have binomial denominators. However, as mentioned before,
that topic is not covered in the BCIT Mathematics 11 Competency Test.
David W. Sabo (2003)
Multiplication and Division With Square Roots
Page 6 of 10
Example 10: Simplify
20 x 7
.
10 x 4
solution:
We can use either of two approaches here. We could start by using property (2) from the
beginning of this document to write
20 x 7
10 x 4

20 x 7
10 x 4
and then simplify the fraction in the square root to get
20 x 7
10 x 4
2

20 x 7
10 x
x3

4
2x 3
 x   2x 

2

x2
2x
 x 2x
as a final, simplified answer. This approach is quite short and efficient.
A second approach is to simplify the square roots first
20 x 7
10 x 4

2 x3
x
5x
x2

2x 5 x
10
10
and then rationalize the denominator:
2x 5 x
10

2x 5 x
10
10
10

2 x 50 x
10
5

x 50 x
5
But
50 x 
52 2 x  5 2 x
and so
x 50 x
5

x  5
5
2x

 x 2x ,
which is exactly the same final answer as we obtained with the first method.
David W. Sabo (2003)
Multiplication and Division With Square Roots
Page 7 of 10
50r 3s 2
Example 11: Simplify
.
5rs
solution:
We’ll use the method of the first approach in Example 10 above:
3 2
50r s
5rs
50

10
3
r2
r
s2
5r s
s

10r 2s
 r 10s
as the final answer.
Example 12: Simplify
18 12t  5 3t .
solution:
Divisions can be written as fractions, so the methods of the last two examples can be used here.
18 12t  5 3t 
18 12t
5 3t
4
18

5

12 t
3t
18
5
4 
18
36
 2 
5
5
as the final answer.
Example 13: Simplify
52 x
x
.
solution:
This is really more of a “rationalize the denominator” problem than it is a dividing problem. We
get
David W. Sabo (2003)
Multiplication and Division With Square Roots
Page 8 of 10
52 x
x
x

x
5  2 x  x   5  2
 x  x 
x

x

x
5 x  2x
x
as the final, simplest result.
35p  7q .
Example 14: Simplify
solution:
5
35 p
35 p  7q 
35 p

7q

7q
5p
q
q

5p
q

5pq
q
q
as the final simplified result with the denominator rationalized.
Example 15: Expand and simplify the product:
x  4 
6
 x  4 
6
.
solution:
The final result here may be a bit of a surprise. Basically, we are asked to multiply one trinomial
by another, and simplify the result. The multiplication step is a bit tedious, but you are well
familiar with the method:
x  4 
6
 x  4 

 x x  4 
6



6  (4) x  4 

  x  x    x  4    x   6
David W. Sabo (2003)
 
6 
6
 x  4 
6


Multiplication and Division With Square Roots
Page 9 of 10

  4  x    4  4    4   6


 x 2  4x 
x  
6
6
  4  

6
6 x  4x  16  4 6 
 
6

6 x 4 6 6
 x 2  8 x  10
as the final simplified answer. Notice that all of the square root terms in the second last line have
cancelled out, because they each occur in pairs of opposite sign.
If you look back at Example 3 in the document on factoring trinomials, you’ll see that we
attempted to factor the trinomial x2 + 8x + 10 into a product of two binomials there, and failed in
the attempt. This example here shows that x2 + 8x + 10 can be “factored” into a product, but it is
a product of two trinomials (hardly a simplification!) and those trinomials involve the square root of
a number. You can see that the systematic trial and verification method we used to factor
trinomials into products of binomials will not work to get this factorization.
David W. Sabo (2003)
Multiplication and Division With Square Roots
Page 10 of 10