Section 9-5 Factoring TrinomialsSPI 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) = x2 + 8x + 15

• Quadratic equation: the highest exponent is 2; (i.e. x2)

Factoring a Quadratic Equationx2 + bx + c (b and c are constants) into Two Binomials

Factor x2 + 8x + 15. (x + 3)(x + 5)

How do you get the first term (x2)?How do you get the second term (8x)?How do you get the third term (15)?

Factors of 15 Sum of Factors1 and 15 163 and 5 8

x2 + 8x + 15 = (x + 3)(x + 5).

Find the factors of 15. Identify the pair that has a sum of 8.

x ∙ x5x + 3x

3 ∙ 5

Factor g2 + 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 g2 - 7g + 10

How does the factors change when you have a negative 2d term?

Factors of 10 Sum of Factors

1 ∙ 10 11

2 ∙ 5 7

Factors of the problem are (g - 2)(g - 5)

Both conditions must be true

Factoring a Quadratic Equation

a. Factor x2 + 13x – 48.

Identify the pair of factors of –48 that has a sum of 13.

b. Factor n2 – 5n – 24.

Identify the pair of factors of –24 that has a sum of –5.

x2 + 13x – 48 = (x + 16)(x – 3)

n2 – 5n – 24 = (n + 3)(n – 8)

Factors of –48 Sum of Factors 1 and –48 –4748 and –1 47 2 and –24 –2224 and –2 22 3 and –16 –1316 and –3 13

Factors of –24 Sum 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

Factor d2 + 17dg – 60g2. Factors of –60 Sum of Factors 1 and –60 –5960 and –1 59 2 and –30 –2830 and –2 28 3 and –20 –1720 and –3 17

Find the factors of –60.

Identify the pair that has a sum of 17.

d2 + 17dg – 60g2 = (d – 3g)(d + 20g)

What happens when there are 2 variables?

Factor h2 – 4hk – 77k2. Factors of –77 Sum of Factors 1 and –77 –7677 and –1 7611 and –7 4 7 and –11 - 4

h2 – 4hk – 77k2 = (h + 7k)(h – 11k)