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THE “QF” METHOD FOR FACTORING QUADRATICS
Remember our old friend
f x
( )
=
4
x
3
−
5
x
2
−
7
x
+
2
.
Let’s say we want to factor this completely over
C
.
From
Part B
on the Rational Zero Test, we found a list of candidates for rational zeros.
It turned out that 2 was, in fact, a zero. Therefore, by the Factor Theorem,
x
−
2
( )
was a
factor of
f x
( )
. After performing Synthetic Division, we found that:
4
x
3
−
5
x
2
−
7
x
+
2
=
x
−
2
( )
⋅
4
x
2
+
3
x
−
1
( )
Trialanderror can be used to factor the quadratic factor, but this method makes some
people nervous. There is a more systematic alternative offered to us by the Quadratic
Formula (QF). Ordinarily, we factor before finding zeros, but we will reverse that here.
Use the QF to find the zeros of
4
x
2
+
3
x
−
1
; in other words, solve
4
x
2
+
3
x
−
1
=
0
.
Observe:
a
=
4
,
b
=
3
, and
c
=
−
1
.
x
=
−
b
±
b
2
−
4
ac
2
a
=
−
3
( )
±
3
( )
2
−
4 4
( )
−
1
( )
2 4
( )
=
−
3
±
25
8
=
−
3
±
5
8
x
=
−
3
+
5
8
=
2
8
=
1
4
or
x
=
−
3
−
5
8
=
−
8
8
=
−
1
The zeros of
4
x
2
+
3
x
−
1
are
1
4
and
−
1
.
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View Full DocumentRemember that the leading coefficient of
4
x
2
+
3
x
−
1
was
a
n
=
4
.
An LFT Form of
4
x
2
+
3
x
−
1
is, therefore:
4
x
2
+
3
x
−
1
=
4
x
−
1
4
⎛
⎝
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 Fall '11
 staff
 Calculus, Factoring

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