mathematics 8 FACTORING-POLYNOMIALS-M1w1.pptx

math 8

1. FACTORING POLYNOMIALS Math Teacher

2. Lesson 1: Factoring with common monomial factor Lesson 2: Factoring difference of two squares Lesson 3: Factoring the Sum and Difference of Two Cubes

3. OBJECTIVES: determine patterns in factoring polynomials; and 2. factor polynomials completely and accurately using the greatest common monomial factor (GCMF); 3. factor the difference of two squares; and 4. factor the sum and difference of two cubes.

4. RECAP: 1. Find the Greatest Common Factor (GCF) of the numbers: 18 and 30 GCF = 6 18 = 1, 2 , 3, 6 , 9, 18 30 = 1, 2 , 3, 5, 6, 10, 15, 30 18 = 1 x 18 = 2 x 9 = 3 x 6 30 = 1 x 30 = 2 x 15 = 3 x 10 = 5 x 6

5. ACTIVITY 1: Find the Greatest Common Factor (GCF) of the numbers: 4 and 10 8 and 12 20 and 24 16 and 48 5 and 25

6. ACTIVITY 1: Find the Greatest Common Factor (GCF) of the numbers: 4 and 10 GCF = 2 4 = 1, 2 , 4 1 = 1, 2 , 5, 4 = 1 x 4 = 2 x 2 10 = 1 x 10 = 2 x 5 10

7. RECAP: 1. Find the Greatest Common Factor (GCF) of the numbers: 18 and 30 GCF = 6 18 = 1, 2 , 3, 6 , 9, 18 30 = 1, 2 , 3, 5, 6, 10, 15, 30 18 = 1 x 18 = 2 x 9 = 3 x 6 30 = 1 x 30 = 2 x 15 = 3 x 10 = 5 x 6

8. RECAP: 2. Find the Greatest Common Factor (GCF) of the numbers: 5 and 25 GCF = 5 5 = 1, 5 25 = 1, 5 , 25 5 = 1 x 5 25 = 1 x 25 = 5 x 5

9. LESSON 1: Factoring by Greatest Common Monomial Factor Example 1. Find the GCF of each pair of monomials. 4𝑥 3 𝑎𝑛𝑑 8𝑥 2 Solution: Step 1. Factor each monomial. 4𝑥 3 =   8𝑥 2 = 2 ∙ 2 ∙ x ∙ x ∙ x 2 ∙ 2 ∙ 2 ∙ x ∙ x

10. Step 2. Identify the common factors. 4𝑥 3 =   8𝑥 2 = 2 ∙ 2 ∙ x ∙ x ∙ x 2 ∙ 2 ∙ 2 ∙ x ∙ x Step 3. Find the product of the common factors. = 4𝑥 2 Hence, 4x 2 is the GCMF of 4x 3 and 8x 2 . 2 ∙ 2 ∙ x ∙ x

11. b. 15𝑦 6 𝑎𝑛𝑑 9𝑧   Step 1. Factor each monomial. 15𝑦 6 =   9𝑧 = Step 2. Identify the common factors. 15𝑦 6 = 9𝑧 = 3 ∙ 5 ∙ y ∙ y ∙ y ∙ y ∙ y ∙ y 3 ∙ 3 ∙ z 3 ∙ 5 ∙ y ∙ y ∙ y ∙ y ∙ y ∙ y 3 ∙ 3 ∙ z

12. Step 3. Find the product of the common factors. Note that 3 is the only common factor.   Hence, 3 is the GCMF of 15𝑦 6 𝑎𝑛𝑑 9𝑧

13. Example 2. Write 6𝑥+ 3𝑥 2 in factored form. Step 1. Determine the number of terms. In the given expression, we have 2 terms: 6𝑥 and 3𝑥 2 .   Step 2. Determine the GCF of the numerical coefficients. COEFFICIENT FACTORS COMMON FACTORS GCF 6 3 1 2 3 6 1 3 1 3 3

14. Determine the GCF of the variables. The GCF of the variables is the one with the least exponent. 𝐺𝐶𝐹 of 𝑥 and 𝑥 2 = 𝑥 Step 3. Find the product of GCF of the numerical coefficient and the variables.   Hence, 3𝑥 is the GCMF of 6𝑥 and 3𝑥 2 . 3 ∙ x = 3x

15. Step 4. Find the other factor, by dividing each term of the polynomial 6𝑥+ 3𝑥 2 by the GCMF 3𝑥. 3x + 3𝑥 2 6x 3x 2 + x Step 5. Write the complete factored form 6𝑥+ 3𝑥 2 = 𝟑𝒙 (𝟐 + 𝒙)

16. LESSON 2: Factoring Difference of Two Squares

17. Example 1: Write 𝑥 2 −9 in completely factored form. Step 1. Express the first and the second terms in exponential form with a power of 2. Step 2: Subtract the two terms in exponential form following the pattern 𝑎 2 − 𝑏 2 . Step 3: Factor completely following the pattern 𝑎 2 − 𝑏 2 = (𝑎 + 𝑏)(𝑎 – 𝑏) Hence, the complete factored form is, 𝑥 2 −9 = (𝑥) 2 − (3) 2 = (𝑥 + 3)(𝑥 – 3). 𝑥 2 = 𝑥 ∙ x = (3) 2 9 = 3 3 ∙ = (𝑥) 2 (𝑥) 2 - (3) 2 (x + 3) (x - 3)

18. Example 2: Write 16𝑎 6 −25𝑏 2 in completely factored form. Step 1. Express the first and the second terms in exponential form with a power of 2. 16𝑎 6 = 4 𝑎 3 4 𝑎 3 ∙ = (4 𝑎 3 ) 2 25𝑏 2 = 5𝑏 ∙ 5𝑏 = ( 5𝑏) 2 Step 2. Subtract the two terms in exponential form following the pattern 𝑎 2 − 𝑏 2 . (4 𝑎 3 ) 2 ( 5𝑏) 2 - Step 3: Factor completely following the pattern 𝑎 2 − 𝑏 2 = (𝑎 + 𝑏)(𝑎 – 𝑏) (4 𝑎 2 + 5b) (4 𝑎 2 - 5b)

19. Hence, the complete factored form of 16𝑎 6 −25𝑏 2 is, 16𝑎 6 −25𝑏 2 = (4𝑎 3 ) 2 − (5𝑏) 2 = ( 4𝑎 3 + 5𝑏) ( 4𝑎 3 − 5𝑏)

20. PATTERN: LESSON 3 : Factoring Sum and Difference of Two Cubes

21. Example 1: Factor 𝑦 3 – 27. Look for the two terms 𝑎 and 𝑏 by expressing every term to the power of 3. 𝑦 3 = (𝑦) 3 27 = (3) 3 Using the pattern, 𝑎 = 𝑦 and 𝑏 = 3 . By substituting to 𝑎 3 − 𝑏 3 = ( 𝑎 − 𝑏)( 𝑎 2 + ab + 𝑏 2 ): 𝑦 3 – 27 = 𝑦 3 - 3 3 = (y - 3) ( 𝑦 2 + 3𝑦 + 3 2 ) = (y - 3) ( 𝑦 2 + 3𝑦 + 9 )

22. Example 2: Factor 1 + 8𝑘 3 Look for the two terms 𝑎 and 𝑏 by expressing every term to the power of 3. 1 = (1) 3 8𝑘 3 = (2k) 3 So, 𝑎 = 1 and 𝑏 = 2𝑘 , substituting it to 𝑎 3 + 𝑏 3 =( 𝑎+𝑏)( 𝑎 2 −ab+𝑏 2 ), we have: 1 + 8𝑘 3 = (1 + 2k) [ ( 1) 2 - ( 1) + (2k) 2 ] = ( 1 - 2k (2k ) + 4k 2 ) (1 2k) +