Transcript Factoring Trinomials of the Type x2 + bx + c

Lesson 8-5 Warm-Up
ALGEBRA 1
“Factoring Trinomials of the
Type x2 + bx +c” (8-5)
What is a
“trinomial”?
Trinomial: a polynomial that consists of three unlike terms
Examples:
x2 + 7x + 12
x2 + bx + c
How do you
factor a
trinomial?
To factor a trinomial of the form x2 + bx + c, you must find two numbers that have
a sum of b and a product of c
Example: Factor x2 + 7x + 12
Notice that the coefficient of the middle term, b or 7, is the sum of 3 and 4.
Also, the constant, c or 12, is the product of 3 and 4. Therefore, you can now
create two binomials whose product is x 2 + 7x + 12.
x 2 + 7x + 12. = (x +3)(x + 4)
Check: Does (x +3)(x + 4) = x 2 + 7x + 12?
(x +3)(x + 4) = x 2 + 4x + 3x + 12
= x 2 + 7x + 12 
FOIL
Combine like terms.
ALGEBRA 1
“Factoring Trinomials of the
Type x2 + bx +c” (8-5)
How do you find
two numbers that
have a sum of b
and a product of
c?
Method 1: Create a Table: Title one column “Factors of (Constant)” or
“Factors of “c” and the other column “Sum of the Factors”. Then, fill in the
table with the number pairs that are factors of the constant.
Example: Factor x2 + 7x + 12
To factor this polynomial, we’ll need to find factors pairs of 12 (two numbers
whose product is 12) whose sum is 7. To do this create a table.
ALGEBRA 1
“Factoring Trinomials of the
Type x2 + bx +c” (8-5)
Method 2: Use an Area Model in Reverse: Arrange the Algebra Tiles that model
the trinomial into a rectangle. The sides of the rectangle (length and width) are the
factors of the trinomial. Tip: Think about how to end with the number of desired
“1” tiles.
Example: Factor x2 + 7x + 12
n n n n
x
x
x
n
x xx x
1 1 1 1
1 1 1 1
1 1 1 1
x+ 4
3n + 1
x2
x+ 3
3n + 1
x+ 3
x+ 4
n
2n + 7
ALGEBRA 1
“Factoring Trinomials of the
Type x2 + bx +c” (8-5)
Example: Factor d2 – 17d + 42
To factor this polynomial, we’ll need to find factors pairs of 42 (two numbers
whose product is 42) whose sum is -17. To do this create a table.
So, d2 - 17x + 42 = (d - 3)(d - 14)
Check: Does (d -3)(d - 14) = d 2 - 17x + 42?
(d -3)(d - 14) = d 2 – 3d – 14d + 42
= d2 – 17d + 12 
FOIL
Combine like terms.
ALGEBRA 1
Factoring Trinomials of the Type x2 + bx + c
LESSON 8-5
Additional Examples
Factor x2 + 8x + 15.
Find the factors of 15. Identify the pair that has a sum of 8.
Factors of 15
1 and 15
3 and 5
Sum of Factors
16
8
x2 + 8x + 15 = (x + 3)(x + 5).
Check: x2 + 8x + 15
(x + 3)(x + 5)
= x2 + 5x + 3x + 15
= x2 + 8x + 15
ALGEBRA 1
Factoring Trinomials of the Type x2 + bx + c
LESSON 8-5
Additional Examples
Factor c2 – 9c + 20.
Since the middle term is negative, find negative factors of 20 (a negative times
a negative equals a positive).
Identify the pair that has a sum of –9.
Factors of 20
–1 and –20
–2 and –10
–4 and –5
Sum of Factors
–21
–12
–9
c2 – 9c + 20 = (c – 5)(c – 4)
ALGEBRA 1
Factoring Trinomials of the Type x2 + bx + c
LESSON 8-5
Additional Examples
a. Factor x2 + 13x – 48.
Identify the pair of factors of –48
that has a sum of 13.
Factors of –48
1 and –48
48 and
–1
2 and –24
24 and
–2
3 and –16
16 and
–3
Sum of Factors
–47
47
–22
22
–13
13
x2 + 13x – 48 = (x + 16)(x – 3)
b. Factor n2 – 5n – 24.
Identify the pair of factors of –24 that
has a sum of –5.
Factors of –24
1 and –24
24 and –1
2 and –12
12 and –2
3 and –8
Sum of Factors
–23
23
–10
10
–5
n2 – 5n – 24 = (n + 3)(n – 8)
ALGEBRA 1
Factoring Trinomials of the Type x2 + bx + c
LESSON 8-5
Additional Examples
Factor d 2 + 17dg – 60g 2 .
Find the factors of –60.
Identify the pair that has a sum of 17.
Factors of –60
1 and –60
60 and
–1
2 and –30
30 and
–2
3 and –20
20 and
–3
Sum of Factors
–59
59
–28
28
–17
17
d2 + 17dg – 60g2 = (d – 3g)(d + 20g)
ALGEBRA 1
Factoring Trinomials of the Type x2 + bx + c
LESSON 8-5
Lesson Quiz
Factor each expression.
1. c2 + 6c + 9
(c + 3)(c + 3)
4. y2 + y – 110
(y + 11)(y – 10)
2. x2 – 11x + 18
(x – 2)(x – 9)
3. g2 – 2g – 24
(g – 6)(g + 4)
5. m2 – 2mn + n2
(m – n)(m – n)
ALGEBRA 1