Transcript Lesson 4 - Hays High School

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
Use a quadratic equation to find two real numbers
that have a sum of 5 and a product of –14.
A. 10, –4
B. 5, –1
C. –2, 7
D. –5, 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(5a – 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 = 32
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(x + 4) = 0
(x + 6)(5x + 4) = 0
x+6=0
x = –6
5x + 4 = 0
Factor the GCF from each
grouping.
Distributive Property
Zero Product Property
Solve each equation.
Factor Trinomials
Answer:
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)
Solve Equations by Factoring
ARCHITECTURE The entrance to an office
building is an arch in the shape of a parabola
whose vertex is the height of the arch. The height
of the arch is given by h = 9 – x 2, where x is the
horizontal distance from the center of the arch.
Both h and x are measured in feet. How wide is the
arch at ground level?
To find the width of the arch at ground level, find the
distance between the two zeros.
Solve Equations by Factoring
9 – x2 = 0
Original expression
x2 – 9 = 0
Multiply both sides by –1.
(x + 3)(x – 3) = 0
x + 3 = 0 or x – 3 = 0
x = –3
x= 3
Difference of squares
Zero Product Property
Solve.
Answer: The distance between 3 and – 3 is
3 – (–3) or 6 feet.
Solve Equations by Factoring
Check
9 – x2 = 0
?
9 – (3)2 = 0
?
9–9= 0
0= 0 
or
?
9 – (–3)2 = 0
?
9–9 =0
0 =0 
TENNIS During a match, Andre hit a lob right off
the court with the ball traveling in the shape of a
parabola whose vertex was the height of the shot.
The height of the shot is given by h = 49 – x 2, where
x is the horizontal distance from the center of the
shot. Both h and x are measured in feet. How far
was the lob hit?
A. 7 feet
B. 11 feet
C. 14 feet
D. 25 feet