2.2 – Simplifying Rational Expressions
In general you want to express both the numerator and denominator as factors so that any
common factors can be divided out to create a simplified equivalent expression. It is basically the
same as simplifying a rational number (fraction) with the added step of having to state restrictions
on any variables that might have been cancelled out. Restrictions exist because one can not divide
by zero, and so any variable occurring in the denominator is restricted from value(s) that would
cause this to occur.
Example 1:
a)
The common
factor (4)
divides out
Cancel out common factors to simplify the following expression.
b)
12
8
( 4)(3)
=
( 4)( 2)
3
=
2
4x 2
8x
( 4)( x )( x )
=
( 4)( 2)( x )
x
= , x≠0
2
a)
Set factored
denominator
equal to zero
to find
restrictions.
This is easiest
to do in
factored form
before terms
cancel out.
x–2=0
∴ x=2
d)
2x 2 − 4x
8x
2 x ( x − 2)
=
2 x ( 4)
x−2
=
, x≠0
4
We need to state restrictions
to inform anyone using the
simplified version of our
expression that certain values
are not allowed. This insures
both expressions are truly
equivalent.
restriction
(4x) factor
cancels out
Example 2:
c)
Simplify the following expressions and state any restrictions.
6
6 x − 12
6
=
6( x − 2)
1
=
, x≠2
( x − 2)
3 − 2x
4x − 6
− 2x + 3
=
2( 2 x − 3)
− 1( 2 x − 3)
=
2 ( 2 x − 2)
−1
3
=
, x≠
2
2
b)
4x2 y3
20 x 3 y
4 xxyyy
=
4(5) xxxy
=
e)
y2
,
5x
c)
x≠0& y≠0
x 2 + 3 x − 15
x 2 + 8 x + 15
( x + 5)( x − 3)
=
( x + 5)( x + 3)
x−3
=
, x ≠ −5,−3
x+3
24 x 3 + 8 x 2 − 12 x
4x
2
4 x (6 x + 2 x + 3)
=
4x
= 6 x 2 + 2 x + 3, x ≠ 0
Both x & y
variables have
restrictions
f)
4 x 2 y + 10 xy
6 xy 2 − 8 x 2 y
2 xy ( 2 x + 5)
=
2 xy (3 y − 4 x )
2x + 5
=
, x ≠ 0,
3y − 4x
Any combination
of x & y in this
ratio would give
zero denominator
2.2 – simplifying rational expressions
y ≠ 0,
y≠
4
x
3
2.2 – Simplifying Rational Expressions Practice Questions
1. Simplify the following expressions and state any restrictions.
a)
− 6 x2 y3
30 x 3 y
b)
d)
2x − 1
4 − 8x
e)
g)
t 2 − 8t + 15
t 2 − 25
h)
2. Show
x +1
is equivalent to;
x −1
( y − 1)( y + 2)
( y + 3)( y − 1)
x2 − 1
1 − x2
6 x 2 + 11x + 3
6 x 2 + 13 x + 6
a) −
1+ x
1− x
c)
y−2
y − 5y + 6
f)
a 2 + 8a + 16
a 2 − 16
i)
2 x 2 − xy − y 2
4 x 2 − 4 xy − 3 y 2
b)
x2 + x
x2 − x
2
3. Write a rational expression in one variable so that the restrictions are;
a) x ≠ 2
b) x ≠ −1,0
c) y ≠
1 3
,−
2 4
d) x ≠ ± 5
4. Find the ratio of volume to surface area for a rectangular prism with dimensions of (x + 4) by
(x + 4) by (2x – 2).
Answers 1.a)
f)
y+2
1
1
1
− y2
, y ≠ −3,1 c)
, y ≠ 2,3 d) − , x ≠
, x, y ≠ 0 b)
e) −1, x ≠ ±1
y +3
y −3
4
2
5x
x− y
1 3
a+4
t −3
3x + 1
2 3
, x ≠ − y, y
, a ≠ ±4 g)
, t ≠ ±5 h)
, x ≠ − ,− i)
3x + 2
3 2
2x − 3y
2 2
a−4
t +5
3. answers may vary from these a)
4.
1
1
1
1
b)
c)
d) 2
x−2
x( x − 2)
(2 y − 1)(4 y + 3)
x −5
( x + 4)( x − 1)
, x > 1 because measurements have to be positive
5x
2.2 – simplifying rational expressions