Algebra 1 Notes SOL A.2 Type II, III Factoring
Mrs. Grieser
Name: _____________________________________ Date: _____________ Block: _________
Factoring Binomial Squares
Review Type I Factoring: Factor out greatest common monomial factor
•
Factor: 2x2 + 4x ____________________________
Review Special Products: (a + b)(a – b) = a2 – b2
•
Multiply: (x + 3)(x – 3) __________________ (2x - 5)(2x + 5) _________________
•
Word backwards: What factors produce x2 – 16? _________________________
Type II Factoring – Two Terms that are Perfect Squares
•
If a binomial has terms that are the difference of perfect squares, use the sum and
difference special product to easily factor the binomial.
Examples: Factor the binomials below
a) y2 – 16
•
•
c) 4x2 - 25
b) x2 - 49
d) x4 – 16 (Advanced!)
Sometimes there is a common factor in a polynomial.
IMPORTANT!!! Factor common factors out first (type I factoring); then factor the perfect
square that remains.
Examples: Factor the binomials below (factor out the GCF first!)
a) 3x2 - 27
b) 4x2 – 16
c) 8x2 - 50
You try: Factor the binomials below.
a) m2 – 121
b) 9n2 – 64
c) 3y2 – 147z2
d) 4x2 - 400
e) 72 – 32y2
f) 48x2 – 12y2
Type III Factoring – Trinomials with Leading Coefficient 1 (form: x2 + bx + c)
•
To factor trinomials of the form x2 + bx + c, you must ask the question:
“What do you multiply to get the last number (c), and add to get the middle number (b)?”
•
What two numbers multiply to the top number and add to the bottom number?
Algebra 1 Notes SOL A.2 Type II, III Factoring
Mrs. Grieser Page 2
STEPS: Factor x2 + 3x + 2
• Draw two sets of parentheses
(
)(
)
• First terms are both x (why??)
(x
)(x
)
• What factors of 2 add to 3?
• These are the factors!
( x + ____ )(x + ____)
• Verify by multiplying back.
________________________
Examples:
a) x2 + 5x + 6
b) x2 – 6x + 8
Factors are:
(x + 3)(x + 2)
Factors are:
(x
)(x
Verify!
Verify!
c) x2 - x – 2
)
d) x2 + 2x – 15
Factors are:
(x
)(x
)
Verify!
Factors are:
_________________
Verify!
Solving Trinomial Equations (Finding Zeros (Roots))
•
Factor the trinomial, then apply the zero product property to find the solution(s).
a) Solve x2 + 8x + 12 = 0
Factor:
b) Solve x2 + 3x = 18
Rewrite: x2 + 3x – 18 = 0
c) Find the zeros (roots) of
f(x) = x2 + x – 20
Set f(x) = 0: x2 + x – 20 = 0
Factor:
Factor:
x = ______________
x = ______________
x = ______________
You try: Factor
a) x2 + 4x + 3
b) x2 – 11x + 24
c) x2 + 6x - 16
Find the roots:
e) x2 + 23x – 24 = 0
f) x2 – 14x = -24
Find the zeros:
g) y = x2 – 12x + 36
h) f(x) = x2 + 3x – 40
d) x2 – 2x - 24