1. Factoring Trinomials of the Type x2 + bx + c
ALGEBRA 1 LESSON 7-5
Factor x2 + 8x + 15.
Find the factors of 15. Identify the pair that has a sum of 8.
Factors of 15 Sum of Factors
1 and
3 and
x2 + 8x + 15 = (x + 3)(x + 5).
= x2 + 5x + 3x + 15
Check: x2 + 8x (x + 3)(x + 5)
= x2 + 8x + 15
7-5

2. Factoring Trinomials of the Type x2 + bx + c
ALGEBRA 1 LESSON 7-5
Factor c2 – 9c + 20.
Since the middle term is negative, find the negative factors of 20.
Identify the pair that has a sum of –9.
Factors of 20 Sum of Factors
–1 and –20 –21
–2 and –10 –12
–4 and –5 –9
c2 – 9c + 20 = (c – 5)(c – 4)
7-5

3. Factoring Trinomials of the Type x2 + bx + c
ALGEBRA 1 LESSON 7-5
a. Factor x2 + 13x – 48.
b. Factor n2 – 5n – 24.
Identify the pair of factors of –48 that has a sum of 13.
Identify the pair of factors of –24 that has a sum of –5.
Factors of –48 Sum of Factors
1 and –48 –47
48 and –
2 and –24 –22
24 and –
3 and –16 –13
16 and –
Factors of –24 Sum of Factors
1 and –24 –23
24 and –
2 and –12 –10
12 and –
3 and –8 –5
x2 + 13x – 48 = (x + 16)(x – 3)
n2 – 5n – 24 = (n + 3)(n – 8)
7-5

4. Factoring Trinomials of the Type x2 + bx + c
ALGEBRA 1 LESSON 7-5
Factor d2 + 17dg – 60g2.
Factors of –60 Sum of Factors
1 and –60 –59
60 and –
2 and –30 –28
30 and –
3 and –20 –17
20 and –
Find the factors of –60.
Identify the pair that has a sum of 17.
d2 + 17dg – 60g2 = (d – 3g)(d + 20g)
7-5

5. Factoring Trinomials of the Type x2 + bx + c
ALGEBRA 1 LESSON 7-5
Factor each expression.
1. c2 + 6c x2 – 11x g2 – 2g – 24
4. y2 + y – m2 – 2mn + n2
(c + 3)(c + 3)
(x – 2)(x – 9)
(g – 6)(g + 4)
(y + 11)(y – 10)
(m – n)(m – n)
7-5