1. Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
Solve (2x + 3)(x – 4) = 0 by using the Zero Product Property.
(2x + 3)(x – 4) = 0
2x + 3 = or x – 4 = 0
Use the Zero-Product Property.
2x = –3
Solve for x.
x = – 3 2 or
x = 4
Check: Substitute – for x. 3 2
Substitute 4 for x.
(2x + 3)(x – 4) = 0
[2(– ) + 3](– – 4) 0
[2(4) + 3](4 – 4) 0
(0)(– 5 ) = 0 1
(11)(0) = 0
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2. Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
Solve x2 + x – 42 = 0 by factoring.
x2 + x – 42 = 0
(x + 7)(x – 6) = 0
Factor using x2 + x – 42
x + 7 = 0 or x – 6 = 0
Use the Zero-Product Property.
x = –7 or x = 6
Solve for x.
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3. Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
Solve 3x2 – 2x = 21 by factoring.
3x2 – 2x = 21
Subtract 21 from each side.
3x2 – 2x – 21 = 0
(3x + 7)(x – 3) = 0
Factor 3x2 – 2x – 21.
3x + 7 = 0 or x – 3 = 0
Use the Zero-Product Property
3x = –7
Solve for x.
x = – or x = 3 7 3
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4. Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
The diagram shows a pattern for an open-top box. The total area of the sheet of materials used to make the box is 130 in.2. The height of the box is 1 in. Therefore, 1 in.  1 in. squares are cut from each corner. Find the dimensions of the box.
Define: Let x = width of a side of the box.
Then the width of the material = x = x + 2
The length of the material = x = x + 5
Relate: length  width = area of the sheet
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5. Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
(continued)
Write: (x + 2) (x + 5) = 130
x2 + 7x + 10 = 130
Find the product (x + 2) (x + 5).
x2 + 7x – 120 = 0
Subtract 130 from each side.
(x – 8) (x + 15) = 0
Factor x2 + 7x – 120.
x – 8 = 0 or x + 15 = 0
Use the Zero-Product Property.
x = 8 or x = –15
Solve for x.
The only reasonable solution is 8. So the dimensions of the box are 8 in.  11 in.  1 in.
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6. Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
1. Solve (2x – 3)(x + 2) = 0.
Solve by factoring.
2. 6 = a2 – 5a 3. 12x + 4 = –9x2 4. 4y2 = 25
–2, 3 2 – 2 3 5 2
–1, 6
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