2. © William James Calhoun, 2001 Warm up… Find the GCF of 12a 2 b, 90a 2 b 2 c Ashley is planting 120 tomato plants in her garden. In what ways can she arrange them so that she has the same number of plants in each row, at least 5 rows of plants, and at least 5 plants in each row.

3. © William James Calhoun, 2001 9-2: Factoring Using the Distributive Property

4. © William James Calhoun, 2001 OBJECTIVES You must use GCF and distributive tools to factor polynomials and use grouping techniques to factor polynomials with four or more terms.

5. © William James Calhoun, 2001 When we used the distributive property, we turned 4(x + 5) into 4x + 20. Now, we will be using the second stage of factoring to “undistribute” polynomials, turning 4x + 20 into 4(x + 5). To do this, find the GCF of all the terms in the polynomial. Undistribute the GCF from each term and put it in front of a new set of parenthesis. Leave the remainders inside the parenthesis.

6. © William James Calhoun, 2001 EXAMPLE 1: Use the distributive property to factor each polynomial. A. 12mn 2 - 18m 2 n 2 B. 20abc + 15a 2 c - 5ac List the factors of 12mn 2.List the factors of 20abc. List the factors of 18m 2 n 2. 12mn 2 = 2 · 2 · 3 · m · n · n 18m 2 n 2 = 2 · 3 · 3 · m · m · n · n Find the GCF. 23mnxxxxn 6mn 2 Write the 6mn 2, open a set of parenthesis, pull the GCF from the two monomials, write what is left behind, then close the parenthesis. 6mn 2 (2 - 3m) List the factors of 18m 2 n 2. 20abc = 2 · 2 · 5 · a · b · c 15a 2 c = 3 · 5 · a · a · c Find the GCF. 5axxc 5ac Write the 5ac, open a set of parenthesis, pull the GCF from the three monomials, write what is left behind, then close the parenthesis. 5ac(4b + 3a - 1) List the factors of 5ac. 5ac = 5 · a · c

7. © William James Calhoun, 2001 You try… 4d 2 + 16 75b 2 c 3 + 60bc 3

8. © William James Calhoun, 2001 EXAMPLE 3: Factor 12ac + 21ad + 8bc + 14bd. Any time you encounter a polynomial with four terms and are asked to factor it, there will be only one option open to you. Group the first two terms. Group the second two terms. Pull out the GCF of the first two. Make sure the signs of the second pair of terms is the same as the signs of the first pair. If so, write a plus sign. If not, write a minus sign and change the signs of both the second pair. 3a(4c + 7d) Pull out the GCF of the second two. + 2b(4c + 7d) Now, you have two terms with something the same in them… Both terms have a (4c + 7d). Pull the (4c + 7d) out, {{ (4c + 7d) open parenthesis, ( write what is left behind, and close the parenthesis. 3a+2b) (4c + 7d)(3a + 2b) This rule will change in higher mathematics, but for now, you get the easy life. This next process will work for any 4-nomials you encounter.

9. © William James Calhoun, 2001 EXAMPLE 4: Factor 15x - 3xy + 4y - 20. Here is one where the signs will not line up right. You will need to pull out a negative sign from the second pairing. Group the first two terms. Group the second two terms. Pull out the GCF of the first two. Make sure the signs of the second pair of terms is the same as the signs of the first pair. If so, write a plus sign. If not, write a minus sign and change the signs of both the second pair. 3x(5 - y) Pull out the GCF of the second two. + 4(y - 5) Now, you have two terms with something the same in them… Both terms have a (5 - y). Pull the (5 - y) out, open parenthesis, ( write what is left behind, and close the parenthesis. 3x-4) (5 - y)(3x - 4) (5 - y) {{ 3x(5 - y) - 4(-y + 5) 3x(5 - y) - 4(5 - y)

10. © William James Calhoun, 2001 You try… x 2 + 4x + 2x + 8 4b 2 – 12b + 2b - 6

11. © William James Calhoun, 2001 A landscaping company has been commissioned to design a triangular flower bed for a mall entrance. The final dimensions of the flower bed have not been determined, but the company knows that the height will be two feet less than the base. The area of the flower bed can be represented by the equations A = ½ b 2 – b Write this equation in factored form Suppose the base of the flower bed is 16 feet. What will be its area? ½ b 2 = ½ * b* b -b = -1 * b GCF is b b( ½ b – 1) A = ½ b 2 – b = ½ (16) 2 – 16 = ½ (256) – 16 = 128 – 16 = 112

12. © William James Calhoun, 2001 Zero Product Property If the product of two factor is 0, then at least one of the factors must be 0. Symbols: for any real numbers a and b, ab=0, then either a = 0 or b = 0 or both equal zero.

13. © William James Calhoun, 2001 Solve an equation in factored form Solve (x – 2)(4x – 1) = 0 Using the zero product property we know that (x – 2) or (4x – 1) must equal zero. (x – 2)(4x – 1)=0 x – 2 = 0 or 4x – 1= 0 x = 2 4x = 1 x = ¼ The solution set is { 2, ¼ }

14. © William James Calhoun, 2001 Solve 4y = 12y 2 We must first write this in ab = 0 form. 4y = 12y 2 4y – 12y 2 = 0 subtract 12y 2 from both sides 4y(1 – 3y) = 0 factor the GCF of 4y and 12y 2, 4y 4y = 0or 1 – 3y = 0 y = 0 -3y = -1 y = 1/3 Solution set is { 0, 1/3 }

15. © William James Calhoun, 2001 Class Work… Page 484 & 485 # 16 – 36 mult of 4 42, 43, 48, 52, 56