Multiplying fractions is easy: you just multiply the tops, and multiply the bottoms. For instance,
6
7
×
7
11
=
6
×
7
7
×
11
=
42
77
6
7
×
7
11
=
6
×
7
7
×
11
=
42
77
size 12{ { {6} over {7} } times { {7} over {"11"} } = { {6 times 7} over {7 times "11"} } = { {"42"} over {"77"} } } {}
(1)
Now, you may notice that
42774277 size 12{ { {"42"} over {"77"} } } {} can be simplified, since 7 goes into the top and bottom.
4277=42÷777÷7=6114277=42÷777÷7=611 size 12{ { {"42"} over {"77"} } = { {"42" div 7} over {"77" div 7} } = { {6} over {"11"} } } {}. So
42774277 size 12{ { {"42"} over {"77"} } } {} is the correct answer, but
611611 size 12{ { {6} over {"11"} } } {} is also the correct answer (since they are the same number), and it’s a good bit simpler.
In fact, we could have jumped straight to the simplest answer first, and avoided dealing with all those big numbers, if we had noticed that we have a 7 in the numerator and a 7 in the denominator, and cancelled them before we even multiplied!
This is a great timesaver, and you’re also a lot less likely to make mistakes.
If the same number appears on the top and the bottom, you can cancel it before you multiply. This works regardless of whether the numbers appear in the same fraction or different fractions.
But it’s critical to remember that this rule only applies when you are multiplying fractions: not when you are adding, subtracting, or dividing.
As you might guess, all this review of basic fractions is useful because, once again, rational expressions work the same way.
Table 1
3x
2
−
21
x
−
24
x
2
−
16
⋅
x
2
−
6x
+
8
3x
+
3
3x
2
−
21
x
−
24
x
2
−
16
⋅
x
2
−
6x
+
8
3x
+
3
size 12{ { {3x rSup { size 8{2} }  "21"x  "24"} over {x rSup { size 8{2} }  "16"} } cdot { {x rSup { size 8{2} }  6x+8} over {3x+3} } } {}

The problem

=
3
(
x
−
8
)
(
x
+
1
)
(
x
+
4
)
(
x
−
4
)
⋅
(
x
−
2
)
(
x
−
4
)
3
(
x
+
1
)
=
3
(
x
−
8
)
(
x
+
1
)
(
x
+
4
)
(
x
−
4
)
⋅
(
x
−
2
)
(
x
−
4
)
3
(
x
+
1
)
size 12{ {}= { {3 \( x  8 \) \( x+1 \) } over { \( x+4 \) \( x  4 \) } } cdot { { \( x  2 \) \( x  4 \) } over {3 \( x+1 \) } } } {}

Always begin rational expression problems by factoring! Note that for the first element you begin by factoring out the common 3, and then factoring the remaining expression.


When multiplying fractions, you can cancel anything on top with anything on the bottom, even across different fractions

=
(
x
−
8
)
(
x
−
2
)
x
+
4
=
(
x
−
8
)
(
x
−
2
)
x
+
4
size 12{ {}= { { \( x  8 \) \( x  2 \) } over {x+4} } } {}

Now, just see what you’re left with. Note that you could rewrite the top as
x
2
−
10
x
+
16
x
2
−
10
x
+
16
size 12{x rSup { size 8{2} }  "10"x+"16"} {}
but it’s generally easier to work with in factored form.

To divide fractions, you flip the bottom one, and then multiply.
1
2
÷
1
3
=
1
2
⋅
3
=
3
2
1
2
÷
1
3
=
1
2
⋅
3
=
3
2
size 12{ { {1} over {2} } div { {1} over {3} } = { {1} over {2} } cdot 3= { {3} over {2} } } {}
(2)After the “flipping” stage, all the considerations are exactly the same as multiplying.
Table 2
x
2
−
3x
2x
2
−
13
x
+
6
x
3
+
4x
x
2
−
12
x
+
36
x
2
−
3x
2x
2
−
13
x
+
6
x
3
+
4x
x
2
−
12
x
+
36
size 12{ { { { { size 8{x rSup { size 6{2} }  3x} } over {2x rSup { size 6{2} }  "13"x+6} } } over { size 12{ { {x rSup { size 6{3} } +4x} over {x rSup { size 6{2} }  "12"x+"36"} } } } } } {}

This problem could also be written as:
x2−3x2x2−13x+6÷x3+4xx2−12x+36x2−3x2x2−13x+6÷x3+4xx2−12x+36 size 12{ { {x rSup { size 8{2} }  3x} over {2x rSup { size 8{2} }  "13"x+6} } div { {x rSup { size 8{3} } +4x} over {x rSup { size 8{2} }  "12"x+"36"} } } {}. However, the
size 12{ div } {} symbol is rarely seen at this level of math.
12÷412÷4 size 12{"12" div 4} {} is written as
124124 size 12{ { {"12"} over {4} } } {}. 
x
2
−
3x
2x
2
−
13
x
+
6
×
x
2
−
12
x
+
36
x
3
+
4x
x
2
−
3x
2x
2
−
13
x
+
6
×
x
2
−
12
x
+
36
x
3
+
4x
size 12{ { {x rSup { size 8{2} }  3x} over {2x rSup { size 8{2} }  "13"x+6} } times { {x rSup { size 8{2} }  "12"x+"36"} over {x rSup { size 8{3} } +4x} } } {}

Flip the bottom and multiply. From here, it’s a straight multiplication problem. 
=
x
(
x
−
3
)
(
2x
−
1
)
(
x
−
6
)
×
(
x
−
6
)
2
x
(
x
2
+
4
)
=
x
(
x
−
3
)
(
2x
−
1
)
(
x
−
6
)
×
(
x
−
6
)
2
x
(
x
2
+
4
)
size 12{ {}= { {x \( x  3 \) } over { \( 2x  1 \) \( x  6 \) } } times { { \( x  6 \) rSup { size 8{2} } } over {x \( x rSup { size 8{2} } +4 \) } } } {}

Always begin rational expression problems by factoring! Now, cancel a factor of
xx size 12{x} {} and an
x−6x−6 size 12{ left (x  6 right )} {} and you get... 
=
(
x
−
3
)
(
x
−
6
)
(
2x
−
1
)
(
x
2
+
4
)
=
(
x
−
3
)
(
x
−
6
)
(
2x
−
1
)
(
x
2
+
4
)
size 12{ {}= { { \( x  3 \) \( x  6 \) } over { \( 2x  1 \) \( x rSup { size 8{2} } +4 \) } } } {}

That’s as simple as it gets, I’m afraid. But it’s better than what we started with! 
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