1. 8-6 Factoring Trinomials of the Type 𝑎𝑥 2 +𝑏𝑥+𝑐
Hubarth
Algebra

2. When the leading coefficient is greater than one we will use the Grouping Method to
factor the trinomials.
Here is a step by step breakdown of grouping.
Factor 2𝑦 2 +5𝑦+2
First, multiply the first and last terms.
2 x 2 = 4.
2nd, what are the factors of 4 that add to 5.
Note: the sign of the last number
tells you to add or subtract to get
the middle number. 4
the factors are 4 and 1 4
2 2
Third, remove the 5y from the original problem and replace with the factors but now include
the variable. 4y and y
2𝑦 2 +4𝑦+𝑦+2
Fourth, you will group the first two terms and the last two.
(2 𝑦 2 +4𝑦)(𝑦+2)
Fifth, now factor out the common monomial in each group. If there is no common monomial
it will be 1.
2𝑦 𝑦+2 +1(𝑦+2)

3. 2𝑦 𝑦+2 +1(𝑦+2)
You should notice that the value in parenthesis are the same. they should always be the same.
Sixth, you will now bring the common monomial factors together and the values in
parenthesis become one. These are your two binomial factors and your solution.
(𝟐𝒚+𝟏)(𝒚+𝟐)

4. Ex 1 Factor
Factor 6𝑛 2 +23𝑛+7
1. 6 x 7 = 42
2. Find the factors of 42 that add to 23
3. Plug these factors in for the middle term
6𝑛 2 +21𝑛+2𝑛+7 42 42 21 14
4. Group the first two and last two terms
together.
(6𝑛 2 +21𝑛)(2𝑛+7)
2 x 21=42
= 23
5. Now factor each group.
3𝑛 2𝑛+7 +1(2𝑛+7)
6. Group into your factors.
(3𝑛+1)(2𝑛+7)

5. Ex 2 Factoring
Factor 7𝑥 2 −26𝑥−8 56
1. 7 x 8 = 56
2. Because the last term is negative
you subtract to get the middle value. 56 28 14
2 x 28 =56
2 – 28 = -26
3. 7𝑥 2 −28𝑥+2𝑥−8
4. (7𝑥 2 −28𝑥)(+2𝑥−8)
5. 7x(x – 4) +2(x – 4)
6. (7x + 2)(x – 4)

6. Ex 3 Factoring Out a Common Monomial First
Factor 20𝑥 2 +80𝑥+35 completely
What is the GCF for each term?
5 is the GCF for 20, 80 and 35
5( 4𝑥 2 +16𝑥+7) from here we now need to put the 5 aside and factor the
trinomial by grouping.
4 x 7 = 28
(4𝑥 2 +14𝑥)(2𝑥+7) 28
2x(2x + 7)+1(2x + 7) 28 14
2 x 14 = 28
=16
(2x + 1)(2x + 7)
Now, place the factors with the monomial factor
5(2x + 1)(2x + 7)

7. Practice
Factor each expression
1. 2𝑦 2 +5𝑦 𝑛 2 −23𝑛 𝑑 2 −14𝑑−3
4. 2𝑛 2 +𝑛− 𝑦 2 +14𝑦+6
(2y + 1)(y + 2)
(3n – 1)(2n – 7)
(5d + 1)(d – 3)
(2n + 3)(n – 1)
2(2y + 1)(y + 3)