Download Factoring Method Sheet

Factoring Method Sheet
COUNT THE TERMS FIRST
1. Factor by finding the GCF (Greatest Common Factor) ■binomial and trinomial product
1. Find the GCF
2. Divide each term by the GCF –Polynomial ∕Monomial
Example: Factor: 5X3 – 35X2 +10X
5X
5X
5X
Answer: 5X(X2 - 7X +2)
2. Factor by grouping ■ Quadnomial
1. Group the first two terms with parenthesis
2. Find the GCF of each set of terms and factor
3. Write the answer –write the common binomial ONCE and write the two GCF’s as a binomial
4. Check by FOIL
Example: Factor: a2 -3a +2ab – 6b (a2 - 3a) + (2ab - 6b)
a(a - 3) + 2b(a – 3)
Answer:
(a – 3) (a + 2b) or (a + 2b) (a – 3)
Checking:
a2 + 2ab + 2a -6b
2
Example: Factor: 4a + 5ab -10b -8a
(4a2 + 5ab) – (10b – 8a)
a(4a + 5b) - 2(5b - 4a) descending order
a(4a + 5b) -2 (±4a + 5b)
Answer:
(a – 2) (4a + 5b) (check -2 to make sure of sign)
Checking:
4a2 + 5ab – 8a – 10b
3. Factor by Easy Method ■ Trinomial product –GCFfirst, find the GCF and then factor completely
■ Leading coefficient of “1” →(1) x2 + bx + c
■ FOIL form answer
1. Find the signs of the binomials
2. Find the factors of the last term whose sum or difference equals the middle term.
3. Check by FOIL
Example: Factor: x2 + 8x +12 Answer: (x + 2) (x + 6)
Example: x2 – 7x + 12 12÷1=12 12+1=13
Answer: (x - 3) (x - 4)
Checking:x2 – 4x – 3x + 12
Example: x2 – 11x – 12
12÷2=6 6+2=8
12÷3=4 3+4=7
12÷1=12 12±1=11
Answer: (x + 1) (x – 12)
Checking: x2 – 12x +x – 12
4. Factor by Trial Factors (count terms) ■ Trinomial product –GCFfirst, and factor completely Method 3 or 4
■ Leading coefficient is 2 or greater (ax2 + bx + c)
■ FOIL form answer
1. Find the factors of the first and last terms and place them in a chart
2. Do outer and inner FOIL within the chart, the sum or difference must equal the middle term
3. Check by FOIL
Example:
Factor: 2x2 - 7x + 3 Step 1. GCF first and factor completely
1,2
1, 3 Step 2. Chart Step 3. Do outer & inner FOIL with the terms on the outside
3, 1 (1 x 1) + (2 x 3) =7
(add or subtract depending on sign) answer must match
Answer: (x – 3) (2x – 1)
middle term
Checking: 2x2 – 1x – 6x + 3
Step 4. Check by FOIL
Change the trinomial to a quadnomial and factor by grouping. Multiply the first and last terms and find the
factors of the product whose sum or difference equals the middle term. Replace the two factors for the middle
term and factor by grouping.
5. Special Factoring (count terms)
5. The difference of two squares ■ Binomial product
■ Negative last term -GCF first and factor completely by 5A only
1. Both terms must have perfect squares (x 2 – 49 → x, 7)
2. Write the answer → write each term twice and the signs of the binomial will be +, 3. Check by FOIL
Example: Factor: (x 2 – 49) Answer: (x + 7) (x -7)
Example: Factor: 16x2 – 9y2 Answer: (4x + 3y) (4x – 3y) Example: Factor: m4 – 16
(m2 + 4) (m2 - 4) Factor completely
Answer: (m2 + 4) (m+2) (m-2)
6.Perfect square Trinomial ■ Trinomial product
■ Positive last term
-GCF first and factor completely by 3, 4 or 5
1. Both the first and last terms must have perfect squares, find them
2. Multiply the two perfect squares twice (OI of foil) 3x · 2 twice = 6x + 6x
IF the sum equals the middle term write the answer as the square of a binomial (3x + 2) 2
IF the sum does not equal the middle term factor by 3 or 4
Example: Factor: 9x2 + 12 x + 4 step 1 3x + 2 step 2 6x + 6x = middle term Answer: (3x + 2) 2
Example: Factor: x2 + 6x + 9
x+3
3x + 3x =
Answer: (x + 3) 2
2
Example: Factor: 4x + 37 x + 9 step 1 2x + 3 step 2 not= middle term factor by 3 or 4 Answer: (x + 9) (4x + 1)
Principle of zero products
Solve: Multiplication of at least one factor must be “0” to get “0” as a product
■ Standard form ■ Two factors being multiplied AND = to 0
1. Set each factor to 0
2. Solve → Addition property of equations and/or multiplication property of equations
Example: Solve: (x+3) (x-7) =0
X+ 3 = 0 x – 7 = 0
-3 -3
+7 +7
X= -3
x=7
Example: Solve: 2a(3a + 1) = 0
Answer (2): -3, 7
2a = 0 3a + 1 =0
-1 -1
Quadratic equation by factoring
1. The equation must be in standard from
2. Factor: Binomials
Trinomials
Method 1 GCF
Method 3
Method 5 A
Method 4
Method 5B
3. Solve → apply the principle of zero products
Example: Solve: 2x2 + x = 6
2x2 + x - 6 =0
1,2 1, -6
-6, 1
2, -3
-3, 2
(x + 2) (2x -3) =0
x + 2 =0 2x – 3 = 0
-2 -2
+3 +3
-2,
3/2
Answer: -2, 3/2
2a = 0 3a = -1
2
2 3
3
Answer: 0, -1/3