Objectives The student will be able to:

ObjectivesThe student will be able to: Factor using the greatest common factor (GCF) and Factor by Grouping

Review: What is the GCF of 25a2 and 15a? 5a Let’s go one step further… 1) FACTOR 25a2 + 15a. Find the GCF and divide each term by the GCF 25a2 + 15a = 5a( ___ + ___ ) Check your answer by distributing. 5a 3

Find the GCF 6x2 Divide each term by the GCF 18x2 - 12x3 = 6x2( ___ - ___ ) Check your answer by distributing. 2) Factor 18x2 - 12x3. 3 2x

3) Factor 28a2b + 56abc2. GCF = 28ab Divide each term by the GCF 28a2b + 56abc2 = 28ab ( ___ + ___ ) Check your answer by distributing. 28ab(a + 2c2) a 2c2

Factor 20x2 - 24xy • x(20 – 24y) • 2x(10x – 12y) • 4(5x2 – 6xy) • 4x(5x – 6y)

5) Factor 28a2 + 21b - 35b2c2 GCF = 7 Divide each term by the GCF 28a2 + 21b - 35b2c2 = 7 ( ___ + ___ - ____ ) Check your answer by distributing. 7(4a2 + 3b – 5b2c2) 4a2 3b 5b2c2

Factor 16xy2 - 24y2z + 40y2 • 2y2(8x – 12z + 20) • 4y2(4x – 6z + 10) • 8y2(2x - 3z + 5) • 8xy2z(2 – 3 + 5)

Factor by Grouping • When polynomials contain four terms, it is sometimes easier to group like terms in order to factor. • Your goal is to create a common factor. • You can also move terms around in the polynomial to create a common factor. • Practice makes you better in recognizing common factors.

Factor by GroupingExample 1: • FACTOR: 3xy - 21y + 5x – 35 • Factor the first two terms: 3xy - 21y= 3y(x – 7) • Factor the last two terms: + 5x - 35 = 5(x – 7) • The green parentheses are the same so it’s the common factor Now you have a common factor (x - 7) (3y + 5)

Factor by GroupingExample 2: • FACTOR: 6mx – 4m + 3rx – 2r • Factor the first two terms: 6mx – 4m= 2m(3x - 2) • Factor the last two terms: + 3rx – 2r = r(3x - 2) • The green parentheses are the same so it’s the common factor Now you have a common factor (3x - 2) (2m + r)

Factor by GroupingExample 3: • FACTOR: 15x – 3xy + 4y –20 • Factor the first two terms: 15x – 3xy = 3x(5 – y) • Factor the last two terms: + 4y –20 = 4(y – 5) • The green parentheses are opposites so change the sign on the 4 - 4(-y + 5) or – 4 (5 - y) • Now you have a common factor (5 – y)(3x – 4)

ObjectiveThe student will be able to: use the zero product property to solve equations

Zero Product Property If a • b = 0 then a = 0, b = 0, or both a and b equal 0.

1. Solve (x + 3)(x - 5) = 0 Using the Zero Product Property, you know that either x + 3 = 0or x - 5 = 0 Solve each equation. x = -3 or x = 5 {-3, 5}

2. Solve (2a + 4)(a + 7) = 0 2a + 4 = 0 or a + 7 = 0 2a = -4 or a = -7 a = -2 or a = -7 {-7, -2}

3. Solve (3t + 5)(t - 3) = 0 3t + 5 = 0 or t - 3 = 0 3t = -5 or t = 3 t = -5/3 or t = 3 {-5/3, 3}

Solve (y – 3)(2y + 6) = 0 • {-3, 3} • {-3, 6} • {3, 6} • {3, -6}

4 steps for solving a quadratic equation Set = 0 Factor Split/Solve Check • Set the equation equal to 0. • Factor the equation. • Set each part equal to 0 and solve. • Check your answer on the calculator.

4. Solve x2 - 11x = 0 Set = 0 Factor Split/Solve Check GCF = x x(x - 11) = 0 x = 0 or x - 11 = 0 x = 0 or x = 11 {0, 11}

5. Solve. -24a +144 = -a2 Set = 0 Factor Split/Solve Check Put it in descending order. a2 - 24a + 144 = 0 (a - 12)2 = 0 a - 12 = 0 a = 12 {12}

6. Solve 4m2 + 25 = 20m 4m2 - 20m + 25 = 0 (2m - 5)2 = 0 2m - 5 = 0 2m = 5 m = Set = 0 Factor Split/Solve Check

Set = 0 Factor Split/Solve Check 7. Solve x3 + 2x2 = 15x x3 + 2x2 - 15x = 0 x(x2 + 2x - 15) = 0 x(x + 5)(x - 3) = 0 x = 0or x + 5 = 0or x - 3 = 0 {0, -5, 3}

Solve a2 – 3a = 40 • {-8, 5} • {-5, 8} • {-8, -5} • {5, 8}

Solve 4r3 – 16r = 0 • {-16, 4} • {-4, 16} • {0, 2} • {0, 4} • {-2, 0, 2} The degree will tell you how many answers you have!

Find two consecutive integers whose product is 240. Set = 0 Factor Split/Solve Check Let n = 1st integer. Let n + 1 = 2nd integer. n(n + 1) = 240 n2 + n = 240 n2 + n – 240 = 0 (n – 15)(n + 16) = 0

(n – 15)(n + 16) = 0 n – 15 = 0 or n + 16 = 0 n = 15 or n = -16 The consecutive integers are 15, 16or -16, -15.

Objectives The student will be able to: