Solving Quadratic Equations by "ac" Method & Factor by Grouping MathBitsNotebook.com Topical Outline | Algebra 2 Outline | MathBits' Teacher Resources Terms of Use   Contact Person: Donna Roberts

Notice, when we multiply (x + 4)(x + 6) we get x²+ 4x + 6x + 24 which equals x² + 10x + 24.

Whenever you multiply two binomials, you create two "middle" terms, which in many cases are "like" terms and are then combined into one term. It is the creation of these two "middle" terms which becomes the focus of the "ac" Method for factoring trinomials.
Keep in mind that were are working with quadratic expressions of the form ax² + bx + c,

This method of creating two "middle" terms is often called "split the middle" since it endeavors to create those two "middle terms" which will make the factoring process easier. Once the middle is "split" into two parts, the process of "factoring by grouping" is used to arrive at the answer.

Since the method we will be using starts by dealing with the leading coefficient, a,
and the constant term, c, in the quadratic expression ax² + bx + c,
this method is referred to as the "ac" method of factoring.

The "ac" Method of Factoring Quadratics - "Split the Middle":

In plain English: To get "Split the Middle" started, you need to find two numbers that multiply to a • c, and also add to the value of b. These two numbers will be used to form 2 new "middle terms", which will allow for "factoring by grouping" to arrive at the answer.

A set-up for "factoring by grouping" can be seen in the picture above: (x² + 3x) + (5x + 15).
To factor, we would normally need to combine the terms, and then factor the trinomial.
But "factoring by grouping" will be easier:
x(x + 3) + 5(x + 3) factor out the common factor from each binomial
(x + 3)•(x + 5) use Distributive Property in reverse to factor out the shared binomial (x + 3)
DONE!!!

"Split the Middle" Method for Factoring Trinomials ("ac" Method) This "short cut" method gets the "Good Math Keeping" seal of approval. The steps in this process make rational mathematical sense, and the original expression is visibly maintained throughout the solution.

Visual Planner (using the Diamond Method):
You may see a "visual planner" diagram, in the form of a large X, used to organize the numerical values in this "ac" Method of factoring. The two factors in the left and right corners of the X will be the coefficients of the two new "middle terms".

Factor: 8x2 + 26x + 15 In the top of the X, place the product of a = +8 and c = +15. [8 • 15 = 120] In the bottom of the X, place the value of b = +26. Find the factors of 120 that will add up to 26, and place them in the right and left sections of the X. Possible factors of 120 are: 2 • 60       3 • 40       4 • 30       5 • 24       6 • 20       Stop! We found the pair that multiplies to 120 and adds to 26. Insert the two middle terms, using 20 and 6 as coefficients. Group the sections and factor to arrive at the answer. 8x2+ 20x + 6x + 15 (8x2 + 20x) + (6x + 15) 4x(2x + 5) + 3(2x + 5) (2x + 5)(4x + 3)  ANSWER

   Be careful when dealing with negative values in this "ac" Method.

Negative constant term:

1. Factor: 6x2 + 13x - 5
a • c = 6 • (-5) = -30
(+15) • (-2) = -30 (+15) + (-2) = +13 These are the new factors that multiply to a • c (-30) and add to b (+13).
6x2 + 15x - 2x - 5
(6x2 + 15x) - 1(2x + 5) The safest way to "group" when the second middle term is negative is to factor out a "-1" from the second pairing. Less errors are made with this approach.
3x(2x + 5) -1(2x + 5)
(2x + 5)(3x - 1) Answer

Negative middle term:

2. Factor: 2x2 - 13x + 21
a • c = 2 • 21 = 42
(-6) • (-7) = 42 (-6) + (-7) = -13 These are the new factors that multiply to a • c (42) and add to b (-13).
2x2 - 6x - 7x + 21
(2x2- 6x) - 1(7x - 21) Again, the safest way to "group" when the second middle term is negative is to factor out a "-1" from the second pairing. Less errors are made with this approach.
2x(x - 3) - 1•7(x - 3) 2x(x - 3) - 7(x - 3)
(x - 3)(2x - 7) Answer

Solve quadratic equation using "ac" method & factor by grouping.

3. Solve equation: 8x2 + 26x + 15 = 0

Find the factors of a • c that will add to b. a = 8, b = 26, c = 15 8x2 + 20x + 6x + 15 = 0 a • c = 8 • 15 = 120 20 • 6 = 120 (factors of 120 that add to 26) 20 + 6 = 26 8x2 + 20x + 6x + 15 = 0 (create equivalent middle term) (8x2 + 20x) + (6x + 15) = 0 (group) 4x(2x + 5) + 3(2x + 5) = 0 (factor by grouping) (2x + 5)(4x + 3) = 0 2x + 5 = 0       4x + 3 = 0 2x = -5            4x = -3 x - -5/2            x = -3/4

 Mathematically speaking, why does this "short cut" method work?

The reason this method of creating two new "middle terms" works, is due to the relationships between the terms and their coefficients when two binomials are multiplied together.

Consider: (mx + n) • (px + q)
m, n, p, q are constants (numbers)
x is the variable

mpx² +mqx + npx + nq
notice the two middle terms as a result of distribution

mpx² +(mq + np)x + nq
factor out x from the two middle terms

The factored coefficient of the middle term is (mq + np).

The product of the coefficient of the first term and the constant is mp • nq.
NOTE: mp • nq is the product of numbers, so we can rearrange the order (or mq • np).

mq is a factor of mq • np
np is a factor of mq • np
The product of these two factors is mq • np

The terms that create the sum for the middle term (mq and np) are factors of the product of the coefficient of the first term and the constant (mq • np).

(mpx² +mqx) + (npx + nq)
split the middle, and group

mx(px + q) + n(px + q)
factor by grouping

(mx + n) • (px + q)
distributive property in reverse

The results from factoring by grouping return to us that with which we started.

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