﻿ Factor a Trinomial when A is not 1. Using the the AC method.

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# Factor by Grouping : A Formula

How To Factor by Grouping using the AC Formula

We are going to use a method known as the 'ac' method to factor these types of quadratic equations. This is a systematic method that employs factoring by grouping. It is always much easier to look at some example problems before reading generalized steps, but the steps go as follows.

### Steps

If you have a quadratic equation in the form $$\red{a}x^2 + \blue b x + \color{green}{c}$$

• Step 1) Determine the product of $$\blue a \cdot \color{green}{c}$$ (the coefficients in a quadratic equation)
• Step 2) Determine what factors of $$\red{a} \cdot \color{green}{c}$$ sum to $$\blue b$$
• Step 3) ungroup the $$\blue{ middle}$$ term to become the sum of the factors found in step 2
• Step 4) group the pairs.

As I expressed earlier, it's much easier to understand this method by simply walking through a few examples. So don't worry if the steps above seem like algebraic nonsense -- just check out the example problems below.

### Practice Problems

##### Problem 1

$$\text{Factors of 12 and their sum} \\ \begin{array}{c|c | c} \text{Factor 1} & \text{ Factor 2} & Sum \\\hline 12 & 1 & 13 \\\hline 3 & 4 & 7 \\\hline 2 & 6 & \blue 8 \\\hline \end{array}$$

Think of $$\blue 8 x$$ as 2x + 6x
$$3x ^ 2 + \boxed { 2x + 6x} + 4$$

Group the 2 pairs : (3x² + 2x) + (6x + 4)
Remove the common factors: $$x \red{(3x + 2)} + 2\red{(3x + 2)}$$

Rewrite as grouped factors $$(x + 2) \red{(3x + 2)}$$.

##### Problem 2

$$\text{Factors of 12 and their sum} \\ \begin{array}{c|c | c} \text{Factor 1} & \text{ Factor 2} & Sum \\\hline 12 & 1 & 13 \\\hline 3 & 4 & \blue 7 \\\hline 6 & 2 & 8 \\\hline \end{array}$$

Think of $$\blue 7x$$ as 3x + 4x
$$3x^2 + \boxed { 3x + 4x } + 4$$

Group the 2 pairs: (3x² + 3x) + (4x + 4)
Remove the common factors: $$3x \red{(x + 1)} + 4\red{(x + 1)}$$

Rewrite as grouped factors$$(3x + 4)\red{(x + 1)}$$.

##### Problem 3

Use the formula
Product of (a)(c) = (5)(9) = 45
What factors of 45 sum to $$\blue{18}$$?

$$\text{Factors of 45 and their sum} \\ \begin{array}{c|c | c} \text{Factor 1} & \text{ Factor 2} & Sum \\\hline 45 & 1 & 46 \\\hline 9 & 5 & 14 \\\hline 3 & 15 & \blue { 18 } \\\hline \end{array}$$

Think of $$\blue{18}x$$ as 3x + 15x
$$5x^2 + \boxed{3x + 15x} + 9$$
Group the 2 pairs: (5x² + 3x) + (15x + 9)
Remove the common factors: $$x \red{(5x + 3)} + 3 \red{(5x + 3)}$$
Rewrite as grouped factors: $$(x + 3) \red{(5x + 3)}$$

##### Problem 4

Apply our formula
Product of (a)(c) = (2)(3) = 6
What factors of 6 add up to to $$\blue 5$$?

$$\text{Factors of 6 and their sum} \\ \begin{array}{c|c | c} \text{Factor 1} & \text{ Factor 2} & Sum \\\hline 6 & 1 & 7 \\\hline 2 & 3 & 5 \\\hline \end{array}$$

3 & 2

Think of $$\blue 5 x$$ as 2x + 3x
$$2x^2 + \boxed{2x + 3x }+ 3$$
Group the 2 pairs
(2x² + 2x) + (3x + 3)
Remove the common factors:
$$2x \red {(x + 1) }+ 3 \red{ (x + 1)}$$
Rewrite as grouped factors: $$(2x + 3) \red { (x + 1)}$$

##### Problem 5

Remember our formula
Product of (a)(c) = (5)(6) = 30
What factors of 30 add up to to 13?

3 & 10

Think of 13x as 10x + 3x
5x² + 10x + 3x+ 6
Group the 2 pairs :(5x² + 10x) + (3x+ 6)
Remove the common factors: 5x(x + 2) + 3(x + 2)
Rewrite as grouped factors: (5x + 3)(x + 2)

##### Problem 6

You know the deal-- Use our formula for factoring by grouping
Product of (a)(c) = (7)(2) = 14
What factors of 14 add up to to 9?

7 & 2

Think of 9x as 7x + 2x
7x² + 7x + 2x + 2
Group the 2 pairs : (7x² + 7x) + (2x + 2)
Remove the common factors: 7x(x + 1) + 2(x + 1)
Rewrite as grouped factors: (7x + 2)(x + 1)