Chapter section 8.1.1
Topic: quadratic equations
Vocabulary:
• Generic rectangle – is a distribution
box that is filled in.
• Neg times Neg is Pos
Pos times Neg is Neg
Distribution Boxes
Example One
Instructions: Simplify
(2a  6)( a  3)
a -3
2a 2a²
+6 +6a
-6a
-18
 2a  6a  6a  18
2
 2a  18
2
Example Two
Instructions: Solve the generic rectangle puzzle
+a +5
+a +7
+a
+a²
+5a
+a
+a²
+6
+6a +30
+4
+4a +28
+a +3
+7a
+a +9
+a
+a²
+3a
+a
+a²
+9a
+1
+a
+3
-2
-2a
-18
Classwork One
Instructions: Solve the generic rectangle puzzle
+a +3
+a +1
+a
+a²
+3a
+a
+a²
+a
+2
+2a
+6
+5
+5a
+5
+a +7
+a +4
+a
+a²
+7a
+a
+a²
+4a
+9
+9a +63
-6
-6a
-24
Example Three
Instructions: Solve the generic rectangle puzzle
+a -6
+a +7
-a
-a²
+6a
+a
+a²
+6
+6a
-36
+4
+4a +28
+a -4
+7a
+a +5
+a
+a²
-4a
-a
-a²
-5a
-2
-2a
+8
-2
-2a
-10
Classwork Two.
Instructions: Solve the generic rectangle puzzle
+a +4
+a -3
+a
+a²
+4a
+a
+a²
-3a
+8
+8a +32
+3
+3a
-9
+a -6
-a
-2
-a
-a²
+6a
-a
+a²
+2a
-3
-3a
+18
+9
-9a
-18
Example Four
Instructions: Solve the generic rectangle puzzle
+a +4
+2a +2a²
+7
+8a
+7a +28
+2a +3
+a
+2a² +3a
+2
+4a
+6
+a +5
+3a +3a²
+7
+15a
+7a +35
+2a -3
-2a -4a² +6a
-6
-12a +18
Classwork Three
Instructions: Solve the generic rectangle puzzle
+a +5
+3a +3a²
+4
+3a +8
+15a
+a
+3a² +8a
+4a +20
+2
+6a +16
+2a +4
+3a +6a² +12a
+3a +3
+3a 9a² +9a
+4
+4
+8a +16
+12a +12
Example Five
Instructions: Solve the generic rectangle puzzle
-a +6
+2a -2a²
-7
+7a
+12a
-42
-2a -4
+a
-2a²
-3
+6a +12
-4a
-2a +6
+3a -6a² +18a
-5
+10a -30
-2a +8
+3a -6a² +24a
+2
-4a
+16
Classwork Four
Instructions: Solve the generic rectangle puzzle
-a +1
+2a -2a²
-3
+3a
+2a
-3
-2a +1
+3a -6a² +3a
-1
-2a -5
+a
-2a²
-5
+10a +25
-5a
+2a
-1
-a +1
+3a -3a²
+4
-4a
+3a
+4
Example Six
Instructions: Test the generic rectangle puzzle
-2a² +2a
+2a
-2a²
-2
-5a
+10a +25
(+2)(a)(+2)(a) = 4a²
(-2)(a²)(-2) = 4a²
(+10)(a)(-5)(a) = -50a²
(-2)(a²)(+25) = -50a²
Classwork Five
Instructions: Test the generic rectangle puzzle
+3a² +15a
+60a²
+54a²
+4a +20
+6a² +12a
+6a +12
+3a² +9a
+6a +18
+72a²
+108a²
9a²
+9a
+12a +12
Chapter section 8.1.1
Topic: quadratic equations
Vocabulary:
(a  1)( a  2)  0
So
either
a+1 is zero
a  1 0
or
or
a+2 is zero
a20
Example One
Instructions: Factor the quadratic.
a  5a  6  (a __)(a __)
2
 (a  1)( a  6)
-1
+6
test
1 x -6 1 - 6
-1 x 6 6 - 1
Classwork Four
Instructions: Factor the quadratic.
a  7a  8  (a __)(a __)
2
1 and -8
a  6a  40  (a __)(a __)
10 and -4
a  8a  15  (a __)(a __)
-3 and -5
a  8a  16  (a __)(a __)
-4 and -4
2
2
2
Example Two:
• Instructions: Solve
(a  1)( a  6)  0
Has two answers:
If (a-1) is zero
a 1 0
a 1 0
+1
+1
a 1
or
If (a+6) is zero
a6  0
a6  0
-6
-6
a  6
Example Three
Instructions: Solve the quadratic.
a  6a  16  (a __)( a __)
2
(a  2)( a  8)  0
+2
-8
test
a20
-2
a  2
-2
a 8  0
+8
+8
a 8
-2 x 8 8 - 2
2 x -8 2 - 8
Classwork Four
Instructions: Solve the quadratic.
a  12a  18  0
-3 and -6
a  5a  14  0
-2 and 7
a  7a  18  0
2 and -9
2
2
2
a  6a  16  0
2
2 and -8