1. Section 9-5 Factoring Trinomials SPI 23G: select one of the factors of a quadratic equation Objective: Factor Trinomials Prime factors: numbers that are only divisible by 1 and itself Coefficient: constant in front of a variable Multiply Binomials: (x + 3)(x + 5) = x 2 + 8x + 15 Quadratic equation: the highest exponent is 2; (i.e. x 2 )

2. Factoring a Quadratic Equation x 2 + bx + c (b and c are constants) into Two Binomials Factor x 2 + 8x + 15. (x + 3)(x + 5) How do you get the first term (x 2 )? How do you get the second term (8x)? How do you get the third term (15)? Factors of 15Sum of Factors 1 and 1516 3 and 5 8 x 2 + 8x + 15 = (x + 3)(x + 5). Find the factors of 15.Identify the pair that has a sum of 8. x ∙ x 5x + 3x 3 ∙ 5

3. Factor g 2 + 7g + 10 -7-2 ∙ -5 -11-1 ∙ -10 Sum of FactorsFactors of 10 Both conditions must be true Factors of the problem are (g + 2)(g + 5) Factor g 2 - 7g + 10 How does the factors change when you have a negative 2d term? Factors of 10Sum of Factors 1 ∙ 1011 2 ∙ 57 Factors of the problem are (g - 2)(g - 5) Both conditions must be true Factoring a Quadratic Equation

4. a. Factor x 2 + 13x – 48. Identify the pair of factors of –48 that has a sum of 13. b. Factor n 2 – 5n – 24. Identify the pair of factors of –24 that has a sum of –5. x 2 + 13x – 48 = (x + 16)(x – 3) n 2 – 5n – 24 = (n + 3)(n – 8) Factors of –48Sum of Factors 1 and –48–47 48 and –1 47 2 and –24–22 24 and –2 22 3 and –16–13 16 and –3 13 Factors of –24Sum of Factors 1 and –24–23 24 and–1 23 2 and–12–10 12 and–2 10 3 and–8 –5 How does the factors change when the last term is negative? What happens when the 2d and 3d terms are negative? Factoring a Quadratic Equation

5. Factor d 2 + 17dg – 60g 2. Factors of –60Sum of Factors 1 and –60–59 60 and –1 59 2 and –30–28 30 and –2 28 3 and –20–17 20 and –3 17 Find the factors of –60. Identify the pair that has a sum of 17. d 2 + 17dg – 60g 2 = (d – 3g)(d + 20g) What happens when there are 2 variables? Factor h 2 – 4hk – 77k 2. Factors of –77Sum of Factors 1 and –77–76 77 and –1 76 11 and –7 4 7 and –11 - 4 h 2 – 4hk – 77k 2 = (h + 7k)(h – 11k)