(Section 0.7: Factoring Polynomials)
0.7.6
§ Solution
a)
ax
2
+
bx
+
c
=
x
2
+
9
x
+
20
=
x
+
5
()
x
+
4
We want 5 and 4, because they have product =
c
= 20 and (since this is the
a
=
1 case) sum =
b
= 9. We can rearrange the factors:
x
+
4
x
+
5
.
b)
x
2
x
2
±
±
20
x
+
100
10
2
±
Guess that this is a
PST for now.
²³
´´
µ
´´
=
x
±
10
Check:
2
x
±
10
=
±
20
x
²³
´µ
´
2
,o
r
x
±
10
x
±
10
=
x
±
10
2
c)
x
2
±
4
x
±
12
=
x
±
6
x
+
2
How do we know we need
±
6 and +
2?
The constant term,
c
, is negative, so use opposite signs: one "+" and one "
±
."
The middle coefficient,
b
, is negative, so the negative number must be higher in
absolute value than the positive number; it “carries more weight.”
2-factorizations of –12 (which is
c
)
Think: What? • What?? = –12.
Sum =
b
=
±
4?
(
a
=
1 case )
±
12, +1
No
±
6, +2
Yes
±
Can stop
±
4, +3
No
d)
F
+
O
+
I
+
L
=
3
x
2
±
20
x
±
7
=
3
x
+
1
x
±
7
F
= First product (product of the First terms)

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