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Lesson Menu 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 5-Minute Check 1 A.4, –4 B.3, –2 C.2, 0 D.2, –2 Use the related graph of y = x 2 – 4 to determine its solutions.

Over Lesson 4–2 5-Minute Check 2 A.–3, 1 B.–3, 3 C.–1, 3 D.3, 1 Use the related graph of y = –x 2 – 2x + 3 to determine its solutions.

Over Lesson 4–2 5-Minute Check 3 A.0 B.0, between 2 and 3 C.between 1 and 2 D.2, –2 Solve –2x 2 + 5x = 0. If exact roots cannot be found, state the consecutive integers between which the roots are located.

Over Lesson 4–2 5-Minute Check 4 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 5-Minute Check 5 A.zero B.x-intercept C.root D.vertex Which term is not another name for a solution to a quadratic equation?

CCSS 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.

Then/Now You found the greatest common factors of sets of numbers. Write quadratic equations in intercept form. Solve quadratic equations by factoring.

Vocabulary factored form FOIL method

Concept

Example 1 Translate Sentences into Equations (x – p)(x – q)=0Write the pattern. Simplify. Replace p with and q with –5. Use FOIL.

Example 1 Multiply each side by 2 so b and c are integers. Answer: Translate Sentences into Equations

Example 1 A.ans B.ans C.ans D.ans

Concept

Example 2 Factor GCF A. Solve 9y 2 + 3y = 0. 9y 2 + 3y= 0Original equation 3y(3y) + 3y(1) = 0Factor the GCF. 3y(3y + 1)= 0Distributive Property 3y = 0 3y + 1 = 0Zero Product Property y = 0 Solve each equation. Answer:

Example 2 Factor GCF B. Solve 5a 2 – 20a = 0. 5a 2 – 20a= 0Original equation 5a(a) – 5a(4)= 0Factor the GCF. 5a(5a – 4)= 0Distributive Property 5a = 0 a – 4 = 0Zero Product Property a = 0 a = 4Solve each equation. Answer: 0, 4

Example 2 A.3, 12 B.3, –4 C.–3, 0 D.3, 0 Solve 12x – 4x 2 = 0.

Example 3 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 = 0Original equation (x – 3) 2 = 0 Factor using the pattern. x – 3= 0Take the square root of each side. x= 3Add 3 to each side. Answer: 3

Example 3 Perfect Squares and Differences of Squares B. Solve y 2 = 36. y 2 = 32Original equation y 2 – 36= 0Subtract 36 from each side. y 2 – (6) 2 = 0Write in the form a 2 – b 2. (y + 6)(y – 6) = 0Factor the difference of squares. y + 6 = 0y – 6 = 0Zero Product Property y = –6 y = 6Solve each equation. Answer: –6, 6

Example 3 A.8, –8 B.8, 0 C.8 D.–8 Solve x 2 – 16x + 64 = 0.

Example 4 Factor Trinomials A. Solve x 2 – 2x – 15 = 0. ac =–15a = 1, c = –15

Example 4 Factor Trinomials x 2 – 2x – 15= 0Original equation Answer: 5, –3 x 2 + mx + px – 15= 0Write the pattern. x 2 + 3x – 5x – 15= 0m = 3 and p = –5 (x 2 + 3x) – (5x + 15) = 0Group 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.

Example 4 Factor Trinomials B. Solve 5x x + 24 = 0. ac =120a = 5, c = 24

Example 4 Factor Trinomials 5x x + 24= 0Original equation 5x 2 + mx + px + 24= 0Write the pattern. 5x 2 + 4x + 30x + 24= 0m = 4 and p = 30 (5x 2 + 4x) + (30x + 24) = 0Group 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.

Example 4 Factor Trinomials Answer:

Example 4 Solve 6x 2 – 5x – 4 = 0. A. B. C. D.

Example 4 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.

Example 5 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.

Example 5 Solve Equations by Factoring 9 – x 2 =0Original expression x 2 – 9 =0Multiply both sides by –1. (x + 3)(x – 3) =0Difference of squares x + 3 = 0 or x – 3 =0Zero Product Property x = –3 x =3Solve. Answer:The distance between 3 and – 3 is 3 – (–3) or 6 feet.

Example 5 Solve Equations by Factoring Check9 – x 2 =0 9 – (3) 2 =0or9 – (–3) 2 =0 ?? 9 – 9 =09 – 9=0 ?? 0 =00=0

Example 5 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?

End of the Lesson