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 ‘Middle’ or ‘Midpoint’) is the average of two numbers, then there exists a nonnegative number $\,u\,$ such that:

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

Review of Factoring Concepts

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

Note:

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:

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

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!

There are no exercises in this optional section.