# How do you factor by grouping 28s^2 - 37s - 21?

Apr 23, 2015

Factoring $f \left(x\right) = 28 {x}^{2} - 37 x - 21$.

There are 2 methods.

1. Factoring AC method (factoring by grouping) .
Find 2 numbers b1 and b2 that satisfy these 2 conditions:
Sum: $\left(b 1 + b 2\right) = - 37$
and
Product: $\left(b 1 \cdot b 2\right) = a \cdot c = 588$

To find $b 1 \text{ and "b2" }$ compose factor pairs of $a \cdot c = - 588$
Proceed: $\left(1 , - 588\right) , \left(2 , - 294\right) , \ldots . \left(12 , - 49\right) \ldots \text{etc.}$
We find
$b 1 = 12 \text{ and } b 2 = - 49$
( since their sum is $- 37$. )

Next factor by grouping:
$f \left(x\right) = 28 {x}^{2} - 49 x + 12 x - 21$
$= 7 x \cdot \left(4 x - 7\right) + 3 \cdot \left(4 x - 7\right)$

Factored form:
f(x) = (4x - 7)(7x + 3)

2. The new AC method to factor a trinomial f(x)
$f \left(x\right) = 28 {x}^{2} - 37 x - 21. \text{ (1)}$

First convert trinomial (1) to
trinomial: f(x) = x^2 - 37x - 588 " (2),
with $a \cdot c = - 21 \cdot \left(28\right) = - 588$

Compose factor pairs of $a \cdot c = -$
and apply the Rule of Sign for Real Roots.

Proceed: $\left(- 1 , 588\right) \ldots . \left(- 12 , 49\right)$

This sum is $49 - 12 = 37 = - b$

Then b'1 = 12" and b'2 = -49#

Next, divide $b ' 1 \text{ and } b ' 2$ by a
to get $b 1 \text{ and } b 2$ for trinomial (1).

$b 1 = b ' \frac{1}{a} = \frac{12}{28} = \frac{3}{7} ,$
and
$b 2 = b ' \frac{2}{a} = - \frac{49}{28} = - \frac{7}{4.}$

Then, the factored form is:
$f \left(x\right) = \left(x + \frac{3}{7}\right) \left(x - \frac{7}{4}\right)$
$= \left(7 x + 3\right) \left(4 x - 7\right)$

This new AC Method avoids the lengthy factoring by grouping.