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

Answer

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

$$ \red{3}x^2 + \blue 8 x + \color{green}{4} $$

Product of (a)(c) = (3)(4) = 12.
What factors of 12 sum to $$\blue 8 $$?

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

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

$$ \red{3}x^2 + \blue 7 x + \color{green}{4} $$

Product of (a)(c) = (3)(4) = 12.
What factors of 12 sum to $$\blue 7 $$?

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

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

Factor

$$ \red{5}x^2 + \blue{18}x + \color{green}{9} $$

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

$$2x^2 + 5x + 3 $$

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

5x² + 13x + 6

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

7x² + 9x + 2

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)