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ALGEBRA 2:
MODULE 4 LESSON 1
Solving Quadratic Equations by Factoring
Forms of a Quadratic Equation
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Standard Form of a Quadratic Equation
ax² + bx + c = 0, a ≠ 0
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Factored Form of a Quadratic Equation
a(x + p)(x + q) = 0, a ≠ 0
Factoring means to write the terms in multiplication form (as a product).
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Zero Product Property
If ab = 0 then either a = 0 or b = 0 (or both).
The expression must be set equal to zero to use this property.
Zero Product Example: Quadratic in Factored Form (x – 6) (x + 8) = 0
x – 6 = 0 or x + 8 = 0
x = 6 or x = 8
Solve a Quadratic in Factored Form
Solve (x – 11)(x – 1) = 0.
This equation is already in factored form. All we have to do is
use the Zero Product Property and set each factor equal to
zero.
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x – 11 = 0 or x – 1 = 0
x = 11 or x = 1
You may write your solutions as x = {1, 11}. Do not enclose
the values of x in parentheses. They may be misconstrued
as an ordered pair! But brackets may be used to signify a
set of numbers.
Solve by Factoring
1.
2.
3.
Write the quadratic equation in standard form: move all
terms to one side of the equation, usually the left, using
addition or subtraction such that one side of the equation is
equal to zero. Simplify if needed. *To make the process
easier, lets always try to make our quadratic term positive!
[ax² + bx + c = 0]
Write the quadratic equation in factored form: factor the
equation completely. [a(x + p)(x + q) = 0]
Set each factor equal to zero, and solve each equation. [If
ab = 0 then either a = 0 or b = 0]
Solve Quadratics by FactoringExample 1
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Solve x2 - 4x + 3 = 0.
The quadratic equation is in standard form (set
equal to zero). This needs to be factored first.
Factor: x2 - 4x + 3 = (x - 1)(x - 3) *You may want
to use the X factors for this.
(x - 1)(x - 3) = 0
Set each factor equal to zero:
x - 1 = 0 or x - 3 = 0
Solve: x = 1 or x = 3, can also write {1, 3}
Solve Quadratics by FactoringExample 2
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Solve x2 – 8 = 2x.
Write the equation in standard form (set the
equation equal to zero). x2 – 2x – 8 = 0
Factor: x2 – 2x – 8 = (x – 4)(x + 2)
(x – 4)(x + 2) = 0
Set each factor equal to zero:
x – 4 = 0 or x + 2 = 0
Solve: x = 4 or x = -2 or {4, -2}
Solve Quadratics by FactoringExample 3
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Solve x² + 3x = 0
Since it is written in standard form, factor it using GCF: x(x + 3) = 0.
Set all the factors equal to zero, and solve: x(x + 3) = 0
x = 0 or x + 3 = 0
x = 0 or x = –3 OR {0, -3}
Common mistakes on this type of problem:
Do not use X factors- that is for Trinomials.
Do not try to "solve" the equation for "x(x + 3) = 0" by dividing by x. Due to the
fact that the zero product property says either factor could be zero or both dividing
the x assumes that x is not zero. We don’t know that, and you lose one of your
solutions. Therefore, we can't divide by zero…it against math laws!
Even though you are used to factors having variables and numbers (like the factor,
x + 3), a factor might contain only a variable, such as x. Be sure to set this equal to
zero also. If it said 5(x-2)(x+4) = 0, 5 cannot = 0!
Solve Quadratics by FactoringExample 4
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Solve (x + 2)(x + 3) = 20.
It is very common to see this type of problem, and say:
It's already factored! So I'll set the factors equal to 0 and solve.
Remember, the expression must be set equal to zero to use this property.
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Simplify by the distributive property (x + 2)(x + 3) = 20
(x + 2)(x + 3) -20 = 0
x2 + 5x + 6 - 20 = 0
x2 + 5x – 14 = 0 It is in standard form, ready to factor
(x + 7)(x – 2) = 0 Factored, can use X factors
x + 7 = 0 or x – 2 = 0 Set each factor = to 0 and solve
x = –7 or x = 2
Solve Quadratics by FactoringExample 5
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Solve 8x2 = 26x - 15.
Write the equation in standard form (set the
equation equal to zero). 8x2 - 26x + 15 = 0
Factor: (4x – 3)(2x - 5) = 0
-20
-6
Set each factor equal to zero:
4x – 3 = 0 or 2x - 5 = 0
2x
2
4x
8x
Solve: 4x – 3 = 0 or 2x - 5 = 0
-3
-6x
4x = 3
or
2x = 5
x=¾
or
x = 5/2
-5
-20x
15
Back to the Garden
Q: Suppose you are trying to create
a garden. The length of the garden
needs to be six feet longer than the
width. You will be given 40 square
feet of space.
A: We know that the length is (x + 6) and the width is (x). The area is 40 sq. ft. If area equals
length times width, then (x)(x+6) = 40. We want to find the values of that make this equation
true. We can use the method of factoring to solve this problem.
Simplify using the distributive property:
x² + 6x = 40
Write it in standard form:
x² + 6x – 40 = 0
Factor:
(x + 10) (x – 4) = 0
Set each factor equal to zero:
x + 10 = 0 or x – 4 = 0
Solve:
x = -10 or x = 4
Since length cannot be negative, -10 cannot be a solution and the only solution is width = 4. Use 4
to find the length 4 + 6 = 10. The dimensions will be 4 feet x 10 feet.
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