# What is the new AC Method to factor trinomials?

Apr 14, 2015

Use the new AC Method.

#### Explanation:

Case 1. Factoring trinomial type $f \left(x\right) = {x}^{2} + b x + c$.

The factored trinomial will have the form: $f \left(x\right) = \left(x + p\right) \left(x + q\right)$.

The new AC Method finds $2$ numbers $p \mathmr{and} q$ that satisfy these 3 conditions:

1. The product $p \cdot q = a \cdot c$. (When $a = 1$, this product is $c$)
2. The sum $\left(p + q\right) = b$
3. Application of the rule of Signs for real roots.

Reminder of the Rule of Signs.

• When $a \mathmr{and} c$ have different signs, $p \mathmr{and} q$ have opposite signs.
• When $a \mathmr{and} c$ have the same sign, $p \mathmr{and} q$ have the same sign.

New AC Method.

To find $p \mathmr{and} q$, compose factor pairs of $c$, and in the same time, apply the Rule of Signs . The pair whose sum equals to $\left(- b\right)$, or $\left(b\right)$, gives $p \mathmr{and} q$.

Example 1. Factor $f \left(x\right) = {x}^{2} + 31 x + 108.$

Solution. $p \mathmr{and} q$ have the same sign. Compose factor pairs of $c = 108$. Proceed: $\ldots \left(2 , 54\right) , \left(3 , 36\right) , \left(4 , 27\right)$. The last sum is $4 + 27 = 31 = b$. Then, $p = 4 \mathmr{and} q = 27$.
Factoring form: $f \left(x\right) = \left(x + 4\right) \left(x + 27\right)$
.
CASE 2 . Factor trinomial standard type $f \left(x\right) = a {x}^{2} + b x + c$ (1)

Bring back to Case 1.

Convert $f \left(x\right)$ to $f ' \left(x\right) = {x}^{2} + b x + a \cdot c = \left(x + p '\right) \left(x + q '\right)$. Find $p ' \mathmr{and} q '$ by the method mentioned in Case 1 .
Then divide $p ' \mathmr{and} q '$ by $\left(a\right)$ to get $p \mathmr{and} q$ for trinomial (1).

Example 2 . Factor $f \left(x\right) = 8 {x}^{2} + 22 x - 13 = 8 \left(x + p\right) \left(x + q\right)$ (1).

Converted trinomial:

$f ' \left(x\right) = {x}^{2} + 22 x - 104 = \left(x + p '\right) \left(x + q '\right)$ (2).

$p ' \mathmr{and} q '$ have opposite signs. Compose factor pairs of $\left(a c = - 104\right) - \to \ldots \left(- 2 , 52\right) , \left(- 4 , 26\right)$. This last sum is $\left(26 - 4 = 22 = b\right)$. Then, $p ' = - 4 \mathmr{and} q ' = 26$.

Back to the original trinomial (1):

$p = \frac{p '}{a} = - \frac{4}{8} = - \frac{1}{2} \mathmr{and} q = \frac{q '}{a} = \frac{26}{8} = \frac{13}{4}$.

Factoring form

$f \left(x\right) = 8 \left(x - \frac{1}{2}\right) \left(x + \frac{13}{4}\right) = \left(2 x - 1\right) \left(4 x + 13\right) .$

This new AC Method avoids the lengthy factoring by grouping.