This document provides instructions for using the Marley Method to factor quadratic trinomials. It explains that the method relies on systematically finding two factors whose product equals the coefficient of the second term. It gives examples of factoring various quadratic trinomials using this method, emphasizing that the signs of the factors depend on the sign of the third term. It also notes that if a trinomial cannot be factored using this method, the Quadratic Formula must be used instead.

1. Factoring Quadratic Trinomials by the Marley Method
The Marley* Method for Factoring Quadratic Trinomials
The Marley Method relies on a systematic, organized approach to factoring
quadratic trinomials. We assume all common factors have already been
factored out. The method is most useful when the leading term (i.e. x2)
coefficient is not one. Follow on The Marley Method Handout.
First term coefficient is 1. Second term coefficient is 6. Third term is 8. These are important.

2. Multiply the "outside product" and "inside product" with like signs to
determine which arrangement combines to 6 (the coefficient of x which
is the second term).
I circle the ones that work.
I sometimes call the connectors "golden arches" (think McDonalds!!).
Put the factors of "1" in the first position of each set of parentheses and
the factors of 8 in the exact order in the last positions of the parentheses
If nothing works---the polynomial may be prime.
When this happens in the case of an equation, we have to use the
Quadratic Formula which will be part three of this lesson.
Remember that learning to factor will help you be able to solve a problem
like x2 + 6x + 8 = 0 which is an equation for its two solutions.
How?

3. Now we will do a similar problem with the sign in front of the second term positive.
Notice that the work is the same except for the signs.
The signs are the last thing to think about.
If the sign in front of the third term is positive, both signs in the parentheses are the same.
They are both whatever is in front of the second term.
If the sign in front of the third term is negative, the signs are different.
The larger of the "inner" vs. "outer" products gets the sign in front of the second term.
The other factor gets the opposite sign.

4. When the sign in front of the third term (here the 24) is minus, it means that
the signs in the parentheses are different (one plus and one minus). It matters which
sign goes where. There is only one set of correct factors.
To check if you are correct, multiply the two factors back together to get
the original addition.

5. This is a harder problem because the first term coefficient is 10 rather than 1. This
makes the Marley Method superior to others.
Now, mentally multiply the "outside" factors and the "inside" factors trying to get 19.
Start with 2 times 1 and 5 time 6. Can you get a 19 with like signs? No.
Try 2 times 3 and 5 times 2. Can you get a 19 with 6 and 10? No. Try 2 times 6 and 5 times 1.
Can you get 19? No. Try 2 times 2 and 5 times 2. Can you get 19 with like signs? Yes. I have
students circle the factors that worked. Put the factors of 10 (the 2 and 5) in that order in
the first position of each factor and the factors of 6 (3 and 2) in that order in the last position.
Put in the signs according to what we discussed earlier. (Here both signs will be the same—
positive.)Thus: (2x + 3) (5x + 2). You could do this with a permanent marker! No erasing.

6. Multiply the "outside" product to get 2. Inside product is 35. Can you get 3 with a 2 and 35? No.
Try 2(7) and 1(5). Can you get 3 with 14 and 5 with different signs? No.
Now drop down the list of factors of 2 to try 1(1) and 2(35). Can you get 3 with 1 and 70? No.
Try 1(7) and 2(5). Can you get 3 with 7 and 10 with unlike signs? Yes.
So, put the 1 and 2 in the first positions of each binomial factor and 5 and 7 in that order
in the second positions.

7. Check this by using multiplication to see if you get the original problem when you
multiply. It checks. I use FOIL to multiply this mentally and quickly.
That is multiply First terms in each parentheses. X time 2x gives 2x2 Then the outside
product x (-7) gives -7x and add this to the inside product 5(2x) or 10x to give 3x. Then last
Times last or 5(-70 to give -35. This is the 2x2 + 3x – 35 original problem and your are right.

8. Factoring takes some practice. Don’t be discouraged if this seems hard at first.
There are many ways to teach this topic; however, I have taught thousands (really)
of students to factor this way, and it works with practice.
This means now that you can solve the equation x2 + 6x + 8 = 0 by factoring.
One side must be 0 and the other side written in descending order of the power of x.
So we could say by the Marley method: (x + 2)(x + 4) = 0
Then (x + 2) = 0 or (x + 4) = 0
And x = -2 or x = -4 The answers are -2 and -4.
from the previous lesson sent out Tuesday 2/18.
In part three next Tuesday, I will try to do the Quadratic Formula. This is what we have to
use when the polynomial will not factor.
Be patient. This is one of the hardest lessons to learn.