# Application of Averaging Concepts: Factoring a Quadratic

Basic concepts—when deeply understood—can often be applied in seemingly-unrelated and powerful ways.

In this lesson, simple averaging ideas lead to a method of factoring quadratics that is quick, easy, and reliable. No quadratic formula! No guess-and-check! No time spent trying to find ‘numbers that work’!

This section is optional, and is for readers who have already studied factoring. There are no exercises in this section.

Po-Shen Loh discovered this technique. Many thanks to Owen Mortensen for bringing it to my attention.

Learn by example? Jump right to them!

## Review of Averaging Concepts

Let $\,a\,$ and $\,b\,$ be real numbers.

**Definition of average:**
The average of $\,a\,$ and $\,b\,$ is
$\,\frac{a+b}2\,$.
To find the average of two numbers:
find their sum, then divide by $\,2\,$.

**The average of two numbers lies exactly halfway between them:**
See this earlier section
for details on why this is true.

If $\,M\,$ (think ‘**M**iddle’ or
‘**M**idpoint’) is the average of two numbers,
then there exists a nonnegative number $\,u\,$ such that:

- one of the numbers is $\,M + u\,$ (distance $\,u\,$ to the right of $\,M\,$)
- the other number is $\,M - u\,$ (distance $\,u\,$ to the left of $\,M\,$)

$|$ | $|$ | $|$ |

$M - u$ | $M$ | $M + u$ |

## Review of Factoring Concepts

Factor: $\,x^2 + bx + c\,$

Note:

- the coefficient of the $\,x^2\,$ term is $\,1\,$
- the coefficient of the $\,x\,$ term is $\,b\,$
- the constant term is $\,c\,$

**
If you can find these numbers,
then factoring is easy:
**
As per conventional factoring techniques,
to factor $\,x^2 + bx + c\,$
you need two numbers that add to $\,b\,$ and
multiply to $\,c\,$.
For convenience, call these two desired numbers
$\,u_1\,$ (for ‘unknown #1’) and $\,u_2\,$
(for ‘unknown #2’).

Thus, $\,u_1\,$ and $\,u_2\,$ must satisfy:

the sum is $\,b\,$:

$\,u_1 + u_2 = b\,$

the product is $\,c\,$:

$\,u_1u_2 = c\,$

Once you have these numbers, factoring is easy:

$$x^2 + bx + c = (x + u_1)(x + u_2)$$($\bigstar$ Note: There are other methods for factoring that use the relationship between the zeros and factors of polynomials.)

**A conventional-method factoring example:**

Factor: $\,x^2 + \color{red}{5}x + \color{green}{6}\,$

Need: two numbers that add to $\,\color{red}{5}\,$ and multiply to $\,\color{green}{6}\,$

Solution: the numbers $\,\bf\large 2\,$ and $\,\bf\large 3\,$ work ($\,2 + 3 = \color{red}{5}\,$, $\,2\cdot 3 = \color{green}{6}\,$)

The Factorization: $\,x^2 + 5x + 6 = (x + {\bf\large 2})(x + {\bf\large 3})\,$

Note: Here, the numbers are simple enough that the conventional method works well.

**How do you find the desired numbers?**
Often, you find $\,u_1\,$ and $\,u_2\,$ by mentally
checking through various possibilities.

But what if the quadratic has coefficients that are big or
ugly?
What if there are no *real numbers* that work?

The NEW method (discussed next) leads you right to the desired numbers!

## The New Method: ‘Focus First on the Sum’

**Factor:** $\,x^2 + bx + c\,$

**Need:** two numbers that add to $\,b\,$ and
multiply to $\,c\,$.
Call these numbers (the ‘unknowns’)
$\,u_1\,$ and $\,u_2\,$.
Assume that $\,u_1 \ge u_2\,$.
Thus:
$$
\begin{alignat}{2}
u_1 + u_2 &= b &\qquad&\text{(the sum is $\,b\,$)}\cr
u_1u_2 &= c&&\text{(the product is $\,c\,$)}
\end{alignat}
$$

**Find Average of Unknowns:**
*First, focus on the sum information*
(hence the name of this technique).
Use the known sum information to find the average
of the unknowns:
$$\text{average of $\,u_1\,$ and $\,u_2\,$} = \frac{u_1 + u_2}2 = \frac{b}2$$

**
Rename Unknowns,
Knowing They're Centered About Their Average:
**
There exists a nonnegative number $\,u\,$ such that:

- $\,u_1 = \frac b2 + u\,$ (move distance $\,u\,$ to the right of the average)
- $\,u_2 = \frac b2 - u\,$ (move distance $\,u\,$ to the left of the average)

**Solve for $\,u\,$:**
The product of $\,u_1\,$ and $\,u_2\,$ equals $\,c\,$:
$$
u_1\,u_2 = (\ \overbrace{\frac b2 + u}^{u_1}\ )(\ \overbrace{\frac b2 - u}^{u_2}\ ) = c
$$
$$
\begin{alignat}{2}
&\frac{b^2}4 - u^2 = c &\qquad&\text{(FOIL)}\cr
&u^2 = \frac{b^2}4 - c &&\text{(re-arrange terms)}\cr
&u = \pm\sqrt{\frac{b^2}4 - c}&&\text{(take square root)}
\end{alignat}
$$

**Use $\,u\,$ to Find the Unknowns:**
Use the positive square root to get:
$$
\begin{gather}
\color{red}{u_1} \ \ =\ \ \frac b2 + u \ \ =\ \ \frac b2 + \sqrt{\frac{b^2}4 - c}\cr
\color{green}{u_2} \ \ =\ \ \frac b2 - u \ \ =\ \ \frac b2 - \sqrt{\frac{b^2}4 - c}
\end{gather}
$$

Note:
Using either the positive or negative square root
leads to the same two numbers (check this).
Using the *positive* square root is easier
(and always gives $\,u_1 \ge u_2\,$).

**Give the Factorization:**
$$
x^2 + bx + c = (x + \color{red}{u_1})(x + \color{green}{u_2})
$$

In its full generality, the ‘Focus First On The Sum’ method looks complicated—but it truly isn't! A few examples show just how quick, easy, and versatile it is:

## Examples: Factoring Quadratics using the ‘Focus First on the Sum’ Method

- Factors involve only integers
- Factoring when the coefficient of the square term isn't $\,1\,$
- Factors involve irrational numbers
- Factors involve the imaginary number $\,i\,$

## Factors involve only integers

**Factor:** $\,x^2 + 14x + 48\,$

**Need:** two numbers (the ‘unknowns’)
that add to $\,14\,$ and multiply to $\,48\,$

**Find Average of Unknowns:**
$$
\begin{align}
&\text{average of unknowns}\cr
&\qquad = \frac{\text{sum of unknowns}}2\cr
&\qquad = \frac{14}2 = 7
\end{align}
$$

**
Rename Unknowns, Knowing They're Centered
About Their Average:
**
There exists a nonnegative number $\,u\,$
such that the unknowns are: $\,7\pm u\,$

Note: ‘$\,7\pm u\,$’ is a shorthand
for ‘$\,7 + u\,$ or $\,7 - u\,$’.

**Solve for $\,u\,$:**
The product of the unknowns is $\,48\,$:
$$\begin{gather}
(7 + u)(7 - u) = 48\cr
49 - u^2 = 48\cr
u^2 = 1\cr
u = \pm 1
\end{gather}
$$

**Use $\,u\,$ to Find the Unknowns:**
Use $\,u = 1\,$ to get:
$$\begin{gather}
7 + u \ \ =\ \ 7 + 1 \ \ =\ \ \color{red}{8}\cr
7 - u \ \ =\ \ 7 - 1 \ \ =\ \ \color{green}{6}
\end{gather}
$$

**Give the Factorization:**
$$x^2 + 14x + 48 = (x + \color{red}{8})(x + \color{green}{6})$$

## Factoring when the coefficient of the square term isn't $\,1\,$

**Factor:** $\,15x^2 + x - 28\,$

**Need:**
To find the ‘numbers that work’
for a trinomial of the form $\,ax^2 + bx + c\,$,
where $\,a\ne 1\,$, there's a
slight modification.
Now, you must find numbers that multiply to $\,ac\,$
and (still) add to $\,b\,$.

In this example: $\,a = 15\,$, $\,b = 1\,$, $\,c = -28\,$. Thus, we need two numbers that add to $\,1\,$ and multiply to $\,(15)(-28) = -420\,$.

Use the ‘Focus First on the Sum’ method to find the ‘numbers that work’.

**Find Average of Unknowns:**
$$
\begin{align}
&\text{average of unknowns}\cr
&\qquad = \frac{\text{sum of unknowns}}2\cr
&\qquad = \frac{1}2
\end{align}
$$

**
Rename Unknowns, Knowing They're Centered
About Their Average:
**
There exists a nonnegative number $\,u\,$
such that the unknowns are: $\,\frac 12\pm u\,$

**Solve for $\,u\,$:**
The product of the unknowns is $\,-420\,$:
$$\begin{gather}
(\frac 12 + u)(\frac12 - u) = -420\cr\cr
\frac 14 - u^2 = -420\cr\cr
u^2 = \frac 14 + 420 = \frac{1681}4\cr\cr
u = \pm \sqrt{\frac{1681}4} = \pm\frac{\sqrt{1681}}{\sqrt{4}} = \pm\frac{41}2
\end{gather}
$$

**Use $\,u\,$ to Find the Unknowns:**
Use $\displaystyle\,u = \frac{41}2\,$ to get:
$$\begin{gather}
\frac12 + u \ \ =\ \ \frac12 + \frac{41}2 \ \ =\ \ \frac{42}2 \ \ =\ \ 21\cr\cr
\frac12 - u \ \ =\ \ \frac12 - \frac{41}2 \ \ =\ \ -\frac{40}2 \ \ =\ \ -20
\end{gather}
$$

**Give the Factorization:**
Use the factor by grouping
method:
$$
\begin{align}
&15x^2 + x - 28 \cr\cr
&\qquad =\ 15x^2 + 21x - 20x - 28\cr
&\qquad\qquad(\text{use $\,21\,$ and $\,-20\,$ to rewrite middle term})\cr\cr
&\qquad =\ (15x^2 + 21x) - (20x + 28)\cr
&\qquad\qquad(\text{group first two terms and last two terms})\cr\cr
&\qquad =\ 3x(5x + 7) - 4(5x + 7)\cr
&\qquad\qquad\text{(factor each group)}\cr\cr
&\qquad =\ (5x + 7)(3x - 4)\cr
&\qquad\qquad\text{($5x + 7\,$ is a common factor)}
\end{align}
$$
Note: You can't just ‘plop the numbers in place’
when the coefficient of the square term isn't $\,1\,$.

## Factors involve irrational numbers

**Factor:** $\,x^2 - x - 3$

**Need:** numbers that add to $\,-1\,$, multiply to $\,-3\,$

**Find Average of Unknowns:**
$$
\begin{align}
&\text{average of unknowns}\cr
&\qquad = \frac{\text{sum of unknowns}}2\cr
&\qquad = \frac{-1}2
\end{align}
$$

**Rename Unknowns, Knowing They're Centered About Their Average:**
There exists a nonnegative number $\,u\,$ such that the unknowns are: $\,-\frac 12\pm u\,$

**Solve for $\,u\,$:**
The product of the unknowns is $\,-3\,$:
$$\begin{gather}
(-\frac 12 + u)(-\frac12 - u) = -3\cr\cr
\frac 14 - u^2 = -3\cr\cr
u^2 = \frac 14 + 3 = \frac{13}4\cr\cr
u = \pm \sqrt{\frac{13}4} = \pm\frac{\sqrt{13}}2
\end{gather}
$$

**Use $\,u\,$ to Find the Unknowns:**
Use $\displaystyle\,u = \frac{\sqrt{13}}2\,$ to get:
$$\begin{gather}
-\frac12 + u \ \ =\ \ -\frac12 + \frac{\sqrt{13}}2 \ \ =\ \ \frac{-1 + \sqrt{13}}2\cr\cr
-\frac12 - u \ \ =\ \ -\frac12 - \frac{\sqrt{13}}2 \ \ =\ \ \frac{-1 - \sqrt{13}}2
\end{gather}
$$

**Give the Factorization:**
$$
\begin{align}
&x^2 - x - 3\cr\cr
&\qquad =\ \left(x + \frac{-1 + \sqrt{13}}2\right) \left(x + \frac{-1 - \sqrt{13}}2\right)\cr
&\qquad\qquad\text{(plop the numbers in place)}\cr\cr
&\qquad = \ \frac12(2x - 1 + \sqrt{13})\ \frac 12(2x - 1 - \sqrt{13})\cr
&\qquad\qquad\text{(factor $\,\frac 12\,$ out of each group)}\cr\cr
&\qquad =\ \frac14(2x - 1 + \sqrt{13})(2x - 1 - \sqrt{13})
\end{align}
$$

It's satisfying to check this one: $$ \begin{align} &\frac14(\color{purple}{2x} \color{red}{- 1} \color{blue}{+ \sqrt{13}})(2x - 1 - \sqrt{13})\cr &\ = \frac 14\left[\color{purple}{2x}(2x - 1 - \sqrt{13}) \,{\bf\color{red}{-}}\, (2x - 1 - \sqrt{13}) \color{blue}{+ \sqrt{13}}(2x - 1 - \sqrt{13})\right]\cr &\ = \frac14\left[4x^2 - 2x \color{red}{- 2x\sqrt{13}} - 2x + 1 \color{green}{+ \sqrt{13}} \color{red}{+ 2x\sqrt{13}} \color{green}{- \sqrt{13}} - 13\right]\cr &\ = \frac14[4x^2 - 4x - 12]\cr &\ = x^2 - x - 3 \end{align} $$ Hooray!!

## Factors involve the imaginary number $\,i\,$

**Factor:** $\,x^2 + 2x + 3$

**Need:** numbers that add to $\,2\,$, multiply to $\,3\,$

**Find Average of Unknowns:**
$$
\begin{align}
&\text{average of unknowns}\cr
&\qquad = \frac{\text{sum of unknowns}}2\cr
&\qquad = \frac{2}2 = 1
\end{align}
$$

**Rename Unknowns, Knowing They're Centered About Their Average:**
There exists a nonnegative number $\,u\,$ such that the unknowns are: $\,1\pm u\,$

**Solve for $\,u\,$:**
The product of the unknowns is $\,3\,$:
$$\begin{gather}
(1 +u)(1-u) = 3\cr
1 - u^2 = 3\cr
u^2 = -2\cr
u \ \ =\ \ \pm \sqrt{-2} \ \ =\ \ \pm i\sqrt{2}
\end{gather}
$$

**Use $\,u\,$ to Find the Unknowns:**
Use $\,u = i\sqrt{2}\,$ to get:
$$
\begin{gather}
1 + u = 1 + i\sqrt{2}\cr
1 - u = 1 - i\sqrt{2}
\end{gather}
$$

**Give the Factorization:**
$$
x^2 + 2x + 3 = (x + 1 + i\sqrt{2})(x + 1 - i\sqrt{2})
$$

It's also satisfying to check this one. Recall that $\,i^2 = -1\,$: $$ \begin{align} &(x + 1 + i\sqrt{2})(x + 1 - i\sqrt{2}) \cr &\ = x^2 + x \color{red}{- ix\sqrt{2}} + x + 1 \color{green}{- i\sqrt{2}} \color{red}{+ ix\sqrt{2}} \color{green}{+ i\sqrt{2}} - i^2(2)\cr &\ = x^2 + 2x + 1 - (-1)(2)\cr &\ = x^2 + 2x + 1 + 2\cr &\ = x^2 + 2x + 3 \end{align} $$ Hooray again!