Warning: You may distribute the “4” through one of the other factors, but not both!!
We obtain:
4
x
2
+
3
x
−
1
=
4
x
−
1
(
)
x
+
1
(
)
This is the kind of factorization we are typically used to, and it often helps us in
simplification problems.
Example
Simplify
x
+
1
4
x
2
+
3
x
−
1
.
Solution
x
+
1
4
x
2
+
3
x
−
1
=
x
+
1
1
4
x
−
1
(
)
x
+
1
(
)
1
,
x
≠ −
1
=
1
4
x
−
1
,
x
≠ −
1
By the way, we must complete the original problem! We had to factor
f
x
( )
=
4
x
3
−
5
x
2
−
7
x
+
2
over
C
.
Don’t forget the
x
−
2
(
)
factor that we obtained earlier:
4
x
3
−
5
x
2
−
7
x
+
2
=
x
−
2
(
)
4
x
2
+
3
x
−
1
(
)
=
x
−
2
(
)
4
x
−
1
(
)
x
+
1
(
)

2.66
SECTION 2.6: RATIONAL FUNCTIONS
PART A: ASSUMPTIONS
Assume
f
x
( )
is rational and written in the form
f
x
( )
=
N x
( )
D x
( )
,
where
N x
( )
and
D x
( )
are polynomials, and
D x
( )
≠
0
(i.e., the zero polynomial).
Assume
for now
that
N x
( )
and
D x
( )
have no real zeros in common.
Note: The textbook essentially makes this last assumption when it assumes that
N x
( )
and
D x
( )
have no common factors (over
R
) aside from
±
1
, though, in
Part E
, we will
consider what happens when we relax this assumption.
Warning: Even though
x
2
+
x
x
=
x
+
1
∀
x
≠
0
(
)
, we do
not
consider the rational function
[rule]
f
x
( )
=
x
2
+
x
x
to be a polynomial function [rule].
PART B: VERTICAL ASYMPTOTES (VAs)
An asymptote for a graph is a line that the graph approaches.
Example
Let
f
x
( )
=
1
x
−
2
. Find any VAs for the graph of
f
.
Solution
Observe that 1 and
x
−
2
have no real zeros in common.
x
−
2
=
0
⇔
x
=
2
, so the only VA has equation
x
= 2
.