Transcript Document

Five-Minute Check (over Lesson 4–2)
CCSS
Then/Now
New Vocabulary
Key Concept: FOIL Method for Multiplying Binomials
Example 1: Translate Sentences into Equations
Concept Summary: Zero Product Property
Example 2: Factor GCF
Example 3: Perfect Squares and Differences of Squares
Example 4: Factor Trinomials
Example 5: Real-World Example: Solve Equations by Factoring
Over Lesson 4–2
Use the related graph of y = x2 – 4 to determine its
solutions.
A. 4, –4
B. 3, –2
C. 2, 0
D. 2, –2
Over Lesson 4–2
Use the related graph of y = –x2 – 2x + 3 to
determine its solutions.
A. –3, 1
B. –3, 3
C. –1, 3
D. 3, 1
Over Lesson 4–2
Solve –2x2 + 5x = 0. If exact roots cannot be found,
state the consecutive integers between which the
roots are located.
A. 0
B. 0, between 2 and 3
C. between 1 and 2
D. 2, –2
Over Lesson 4–2
Which term is not another name for a solution to a
quadratic equation?
A. zero
B. x-intercept
C. root
D. vertex
Content Standards
A.SSE.2 Use the structure of an expression
to identify ways to rewrite it.
F.IF.8.a Use the process of factoring and
completing the square in a quadratic function
to show zeros, extreme values, and
symmetry of the graph, and interpret these in
terms of a context.
Mathematical Practices
2 Reason Abstractly and quantitatively.
You found the greatest common factors of sets
of numbers.
• Write quadratic equations in intercept form.
• Solve quadratic equations by factoring.
• factored form
• FOIL method
Translate Sentences into Equations
(x – p)(x – q) = 0
Write the pattern.
Replace p with
and q with –5.
Simplify.
Use FOIL.
Translate Sentences into Equations
Multiply each side
by 2 so b and c are
integers.
Answer:
A. ans
B. ans
C. ans
D. ans
Factor GCF
A. Solve 9y 2 + 3y = 0.
9y 2 + 3y = 0
3y(3y) + 3y(1) = 0
3y(3y + 1) = 0
3y = 0 3y + 1 = 0
y=0
Answer:
Original equation
Factor the GCF.
Distributive Property
Zero Product Property
Solve each equation.
Factor GCF
B. Solve 5a2 – 20a = 0.
5a 2 – 20a = 0
5a(a) – 5a(4) = 0
5a(a – 4) = 0
5a = 0
a–4=0
a=0
a=4
Answer: 0, 4
Original equation
Factor the GCF.
Distributive Property
Zero Product Property
Solve each equation.
Solve 12x – 4x2 = 0.
A. 3, 12
B. 3, –4
C. –3, 0
D. 3, 0
Perfect Squares and Differences of Squares
A. Solve x 2 – 6x + 9 = 0.
x 2 = (x)2; 9 = (3)2
First and last terms are
perfect squares.
6x = 2(x)(3)
Middle term equals 2ab.
x 2 – 6x + 9 is a perfect square trinomial.
x 2 + 6x + 9 = 0
(x – 3)2 = 0
x–3 =0
x =3
Answer: 3
Original equation
Factor using the pattern.
Take the square root of each
side.
Add 3 to each side.
Perfect Squares and Differences of Squares
B. Solve y 2 = 36.
y 2 = 36
y2 – 36 = 0
Subtract 36 from each side.
y2 – (6)2 = 0
Write in the form a2 – b2.
(y + 6)(y – 6) = 0
y+6=0
Factor the difference of
squares.
y–6=0
y = –6
Answer: –6, 6
Original equation
y=6
Zero Product Property
Solve each equation.
Solve x 2 – 16x + 64 = 0.
A. 8, –8
B. 8, 0
C. 8
D. –8
Factor Trinomials
A. Solve x 2 – 2x – 15 = 0.
ac = –15
a = 1, c = –15
Factor Trinomials
x 2 – 2x – 15 = 0
Original equation
x2 + mx + px – 15 = 0
Write the pattern.
x 2 + 3x – 5x – 15 = 0
m = 3 and p = –5
(x 2 + 3x) – (5x + 15) = 0
Group terms with
common factors.
x(x + 3) – 5(x + 3) = 0
(x – 5)(x + 3) = 0
x–5=0
x+3 =0
x=5
Answer: 5, –3
x = –3
Factor the GCF from each
grouping.
Distributive Property
Zero Product Property
Solve each equation.
Factor Trinomials
B. Solve 5x 2 + 34x + 24 = 0.
ac = 120
a = 5, c = 24
Factor Trinomials
5x 2 + 34x + 24 = 0
Original equation
5x2 + mx + px + 24 = 0
Write the pattern.
5x 2 + 4x + 30x + 24 = 0
m = 4 and p = 30
(5x 2 + 4x) + (30x + 24) = 0
Group terms with
common factors.
x(5x + 4) + 6(5x + 4) = 0
Factor the GCF from each
grouping.
(x + 6)(5x + 4) = 0
x+6=0
x = –6
5x + 4 = 0
Distributive Property
Zero Product Property
Solve each equation.
Factor Trinomials
Answer:
A. Solve 6x 2 – 5x – 4 = 0.
A.
B.
C.
D.
B. Factor 3s 2 – 11s – 4.
A. (3s + 1)(s – 4)
B. (s + 1)(3s – 4)
C. (3s + 4)(s – 1)
D. (s – 1)(3s + 4)
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