Lesson 5 Menu : Lesson 5 Menu Five-Minute Check (over Lesson 8-4)
Main Ideas and Vocabulary
California Standards
Key Concept: Difference of Squares
Example 1: Factor the Difference of Squares
Example 2: Apply a Factoring Technique More Than Once
Example 3: Apply Several Different Factoring Techniques
Example 4: Standards Example: Solve Equations by Factoring
Example 5: Use Differences of Two Squares
Lesson 5 MI/Vocab : Lesson 5 MI/Vocab Solve equations involving the differences of squares. Factor binomials that are the differences of squares.
Lesson 5 CA : Lesson 5 CA Standard 11.0 Students apply basic factoring techniques to second- and simple third-degree polynomials. These techniques include finding a common factor for all terms in a polynomial, recognizing the difference of two squares, and recognizing perfect squares of binomials.
Standard 14.0 Students solve a quadratic equation by factoring or completing the square. (Key)
Key Concept 8-5 : Key Concept 8-5
Lesson 5 Ex1 : Lesson 5 Ex1 Factor the Difference of Squares A. Factor m2 – 64. m2 – 64 = m2 – 82 Write in the form a2 – b2. = (m + 8)(m – 8) Factor the difference of squares. Answer: (m + 8)(m – 8)
Lesson 5 Ex1 : Lesson 5 Ex1 Factor the Difference of Squares B. Factor 16y2 – 81z2. 16y2 – 81z2 = (4y)2 – (9z)2 16y2 = 4y ● 4y and 81z2 = 9z ● 9z = (4y + 9z)(4y – 9z) Factor the difference of squares. Answer: (4y + 9z)(4y – 9z)
Lesson 5 Ex1 : Lesson 5 Ex1 Factor the Difference of Squares C. Factor 3b3 – 27b. If the terms of a binomial have a common factor, the GCF should be factored out first before trying to apply any other factoring technique. = 3b(b + 3)(b – 3) Factor the difference of squares. 3b3 – 27b = 3b(b2 – 9) The GCF of 3b2 and 27b is 3b.
= 3b[b2 – 32] b2 = b ● b and 9 = 3 ● 3 Answer: 3b(b + 3)(b – 3)
Slide 8 : A
B
C
D A. (b + 3)(b + 3)
B. (b – 3)(b + 1)
C. (b + 3)(b – 3)
D. (b – 3)(b – 3) A. Factor the binomial b2 – 9.
Slide 9 : A
B
C
D A. (5a + 6b)(5a – 6b)
B. (5a + 6b)2
C. (5a – 6b)2
D. 25(a2 – 36b2) B. Factor the binomial 25a2 – 36b2.
Slide 10 : A
B
C
D A. 5x(x2 – 4)
B. (5x2 + 10x)(x – 2)
C. (x + 2)(5x2 – 10x)
D. 5x(x + 2)(x – 2) C. Factor 5x3 – 20x.
Lesson 5 Ex2 : Lesson 5 Ex2 Factor y4 – 625. y4 – 625 Original binomial
= [(y2)2 – 252] y4 = y2 ● y2 and 625 = 25 ● 25
= (y2 + 25)(y2 – 25) Factor the difference of squares.
= (y2 + 25)(y2 – 52) y2 = y ● y and 25 = 5 ● 5 = (y2 + 25)(y + 5)(y – 5) Factor the difference of squares. Apply a Factoring Technique More Than Once Answer: (y2 + 25)(y + 5)(y – 5)
Slide 12 : A
B
C
D A. (y2 + 4)(y2 – 4)
B. (y + 2)(y + 2)(y + 2)(y – 2)
C. (y + 2)(y + 2)(y + 2)(y + 2)
D. (y2 + 4)(y + 2)(y – 2) Factor y4 – 16.
Lesson 5 Ex3 : Lesson 5 Ex3 Factor 6x3 + 30x2 – 24x – 120. 6x3 + 30x2 – 24x – 120 Original polynomial
= 6(x3 + 5x2 – 4x – 20) Factor out the GCF.
= 6[(x3 – 4x) + (5x2 – 20)] Group terms with common factors.
= 6[x(x2 – 4) + 5(x2 – 4)] Factor each grouping.
= 6(x2 – 4)(x + 5) x2 – 4 is the common factor. = 6(x + 2)(x – 2)(x + 5) Factor the difference of squares, x2 – 4, into (x + 2)(x – 2). Apply Several Different Factoring Techniques Answer: 6(x + 2)(x – 2)(x + 5)
Slide 14 : A
B
C
D A. 5(x2 – 9)(x + 5)
B. (5x + 15)(x – 3)(x + 5)
C. 5(x + 3)(x – 3)(x + 5)
D. (5x + 25)(x + 3)(x – 3) Factor 5x3 + 25x2 – 45x – 225.
Lesson 5 Ex4 : Lesson 5 Ex4 Solve Equations by Factoring Replace y with 0. Read the Item Original equation Solve the Item
Lesson 5 Ex4 : Lesson 5 Ex4 Solve Equations by Factoring Zero Product Property Solve each equation. Factor the difference of squares.
Slide 17 : A
B
C
D In the equation m2 – 81 = y, which is a value of m when y = 0?
Lesson 5 Ex5 : Lesson 5 Ex5 Words A is the area of the triangle minus the area of the square that is to be removed. Use Differences of Two Squares
Lesson 5 Ex5 : Lesson 5 Ex5 Translate the verbal statement. Equation A = 64 – x2 Use Differences of Two Squares
Lesson 5 Ex5 : Lesson 5 Ex5 A = 64 – x2 and A0 = 64 64 – x2 = 48 Simplify.
64 – x2 – 48 = 0 Subtract 48 from each side.
16 – x2 = 0 Simplify.
(4 + x)(4 – x) = 0 Factor the difference of squares.
4 + x = 0 or 4 – x = 0 Zero Product Property
x = –4 x = 4 Solve each equation. Answer: Since the length cannot be negative, the only reasonable solution is 4. Use Differences of Two Squares
Slide 21 : A
B
C
D A. 6
B. 9
C. 3
D. –3
End of Lesson 5 : End of Lesson 5