1. Multiply (2n + 3)(n + 4) 2n² + 8n + 3n + 12 2n² + 11n + 12
12-5 Factoring ax² + bx + c
When you multiplied, the _
was ______________ the _
and became _. It was then
____________ . So now we
have to find the correct
___________ for the ____
and ________. The only way 4
Multiply
multiplied times 2
(2n + 3)(n + 4) 8
added to the 3
2n² + 8n + 3n + 12
2n² + 11n + 12
combination
1st
last term
Factor
to factor 2 is _____ but this
must be ________ by the __________ , then ___
___________ together to get the ____________.
( n+ )( n+ )
2&1
add
the products
multiplied
factors of 12
middle term

2. 12-5 Factoring ax² + bx + c
2n² + 11n + 12
combination
for 1st term
combination
for last term
Factor
2 ·
1 ·
3 =
4 = 6 4
(?n + ?)(?n + ?) 10
Not the middle term O
2 ·
1 ·
4 =
3 =
other bin. 8 3 I
one bin. 11
( n + )( n + ) 2 3 1 4

3. 1. 2n² - 3n - 20 2 · 1 · 4 = -5 = 8 -5 ( n + )( n + ) 2 5 1 -4 3 2 ·
Factor
1st term
Combo.
Last term
Combo.
1. 2n² - 3n - 20
2 ·
1 ·
4 =
-5 = 8 -5
( n + )( n + ) 2 5 1 -4 3
No, wrong sign
Outside
2 ·
1 ·
-4 =
5 =
other bin. -8 5
Inside
one bin.
( n + )(n ) 2 5
- 4 -3
2n² - 8n + 5n - 20
2n² - 3n - 20

4. 2. 6y² - 29y - 5 2 · 3 · 1 = -5 = 2 -15 6y² + y - 30y - 5 -13 3 · 2 ·
1st term
Combo.
Last term
Combo.
2 ·
3 ·
1 =
-5 = 2
-15
6y² + y - 30y - 5
-13
No, # too large
check
3 ·
2 ·
1 =
-5 = 3
-10
( y - )( y + ) 1 5 6 1 -7
Even larger
Outside
1 ·
6 ·
1 =
-5 =
other bin. 1
-30
Inside
one bin.
-29

5. 3. Is 2x² + 5x + 3 factorable? If it is, factor
it, if not tell why.
1st term
Last term
2 ·
1 ·
3 =
1 = 6 1
( x )( x + ) 2 3 1 1 7 no
Outside
2 ·
1 ·
1 =
3 =
other bin. 2 3
Inside
one bin. 5

6. 4. 2n² - 3n - 20 = 0 ( n + )(n ) = 0 2 5 - 4 2n + 5 = 0 n -4 = 0 -5 -5
Solve each equation.
4. 2n² - 3n - 20 = 0
1st factor,
Same as # 1
( n + )(n ) = 0 2 5
- 4
Set each
binomial = 0
2n + 5 = 0
n -4 = 0
2n = -5
n = 4
Solve both
equations.
n = -2.5

7. 5. 6y² - 29y -5 = 0 (y - )( y + ) = 0 5 6 1 y - 5 = 0 6y + 1 = 0
1st factor, which
we already did
in problem 2.
(y )( y + ) = 0 5 6 1
y - 5 = 0
6y + 1 = 0
Set each
binomial = 0
6y = -1
y = 5
_ 1 6
y =
Solve both
equations.
6. 12y³ - 58y² - 10y = 0
1) GCF 2y
(6y² - 29y -5) = 0
2) 2 binomials 2y
(y - 5) (6y + 1) = 0
3) Solve
_ 1 6
y = 0, 5,
2y=0
(y - 5)=0
(6y + 1)=0