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

A. 4, –4

B. 3, –2

C. 2, 0

D. 2, –2

Use the related graph of y = x2 – 4 to determine its solutions.

Over Lesson 4–2

A. –3, 1

B. –3, 3

C. –1, 3

D. 3, 1

Use the related graph of y = –x2 – 2x + 3 to determine its solutions.

Over Lesson 4–2

A. 0

B. 0, between 2 and 3

C. between 1 and 2

D. 2, –2

Solve –2x2 + 5x = 0. If exact roots cannot be found, state the consecutive integers between which the roots are located.

Over Lesson 4–2

A. 10, –4

B. 5, –1

C. –2, 7

D. –5, 2

Use a quadratic equation to find two real numbers that have a sum of 5 and a product of –14.

Over Lesson 4–2

A. zero

B. x-intercept

C. root

D. vertex

Which term is not another name for a solution to a quadratic equation?

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.

Simplify.

Replace p with

and q with –5.

Use FOIL.

Multiply each side by 2 so b and c are integers.

Answer:

Translate Sentences into Equations

A. ans

B. ans

C. ans

D. ans

Factor GCF

A. Solve 9y

2 + 3y = 0.

9y

2 + 3y = 0 Original equation

3y(3y) + 3y(1) = 0 Factor the GCF.

3y(3y + 1) = 0 Distributive Property

3y = 0 3y + 1 = 0Zero Product Property

y = 0 Solve each equation.

Answer:

Factor GCF

B. Solve 5a2 – 20a = 0.

5a

2 – 20a = 0 Original equation

5a(a) – 5a(4) = 0 Factor the GCF.

5a(5a – 4) = 0Distributive Property

5a = 0 a – 4 = 0Zero Product Property

a = 0 a = 4Solve each equation.

Answer: 0, 4

A. 3, 12

B. 3, –4

C. –3, 0

D. 3, 0

Solve 12x – 4x2 = 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 Original equation

(x – 3)2 = 0 Factor using the pattern.

x – 3 = 0 Take the square root of each side.

x = 3 Add 3 to each side.

Answer: 3

Perfect Squares and Differences of Squares

B. Solve y

2 = 36.

y

2 = 32 Original equation

y2 – 36 = 0 Subtract 36 from each side.

y2 – (6)2 = 0 Write in the form a2 – b2.

(y + 6)(y – 6) = 0 Factor the difference of squares.

y + 6 = 0 y – 6 = 0 Zero Product Property

y = –6 y = 6 Solve each equation.

Answer: –6, 6

A. 8, –8

B. 8, 0

C. 8

D. –8

Solve x

2 – 16x + 64 = 0.

Factor Trinomials

A. Solve x

2 – 2x – 15 = 0.

ac = –15 a = 1, c = –15

Factor Trinomials

x

2 – 2x – 15 = 0Original equation

Answer: 5, –3

x2 + mx + px – 15 = 0Write 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) = 0Factor the GCF from each grouping.

(x – 5)(x + 3) = 0Distributive Property

x – 5 = 0 x + 3 = 0Zero Product Property

x = 5 x = –3Solve each equation.

Factor Trinomials

B. Solve 5x

2 + 34x + 24 = 0.

ac = 120 a = 5, c = 24

Factor Trinomials

5x

2 + 34x + 24 = 0Original equation5x2 + mx + px + 24 = 0Write 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)= 0Factor the GCF from each grouping.

(x + 6)(5x + 4) = 0Distributive Property

x + 6 = 0 5x + 4 = 0Zero Product Property

x = –6 Solve each equation.

Factor Trinomials

Answer:

Solve 6x

2 – 5x – 4 = 0.

A.

B.

C.

D.

A. (3s + 1)(s – 4)

B. (s + 1)(3s – 4)

C. (3s + 4)(s – 1)

D. (s – 1)(3s + 4)

B. Factor 3s

2 – 11s – 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 – x 2 = 0 Original expression

x 2 – 9 = 0 Multiply both sides by

–1.

(x + 3)(x – 3) = 0 Difference of squaresx + 3 = 0 or x – 3 = 0 Zero Product

Property

x = –3 x = 3 Solve.Answer: The distance between 3 and – 3 is

3 – (–3) or 6 feet.

Solve Equations by Factoring

Check 9 – x

2 = 0

9 – (3)2 = 0 or 9 – (–3)2 = 0? ?

9 – 9 = 0 9 – 9 = 0? ?

0 = 0 0 = 0

A. 7 feet

B. 11 feet

C. 14 feet

D. 25 feet

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?