Factoring Trinomials

Matt Braddock

Abstract

An important topic and essential skill taught in Algebra is factoring trinomials with integer coefficients. It is used frequently in Algebra and courses beyond, and thus it is beneficial for students to learn a single method that is quick to complete. The forthcoming method also treats a leading coefficient of 1 as a special case, rather than presenting students with two separate methods based on the leading coefficient.

How It Works

Consider the trinomial \(2x^{2}-7x-4\). In order to factor this, we must find two numbers that multiply to -8 (\(a\cdot c\)) and add to -7 (\(b\)). Using the diamond/X method to factor, we'd set up the following:
The two numbers that satisfy these conditions are -8 and 1. First, put those two numbers on the sides, then divide each of these two numbers by 2 (\(a\)) and reduce the fractions.
To determine the factors, use both the numerator and denominator of each fraction. The denominator is the coefficient of \(x\), and the numerator is the constant. $$2x^{2}-7x-4=(x-4)(2x+1)$$

A Simple Step-by-Step

Why It Works

If a trinomial \(ax^{2}+bx+c\) can be factored, the product of factors can be written as \((px+q)(rx+s)\) (provided that \(gcf(a,b,c)=1\)). The product expanded gives us \((pr)x^{2}+(ps+qr)x+qs\), which means \(a=pr,\,b=ps+qr,\,c=qs\). We can set up the diamond/X with \(ac\) on top and \(b\) on the bottom:
Conveniently, the \(b\) value gives us our two values. We put them on the sides, divide by \(a\), and reduce.
Thus, the fraction \(\frac{q}{p}\) corresponds to \(px+q\) and the fraction \(\frac{s}{r}\) corresponds to \(rx+s\).

Why Use It

This method is fast (provided students can find the two numbers that satisfy \(ac\) as the product and \(b\) as the sum with relative ease), short (the majority of the work is contained within the diamond/X), and consistent (it works for all integer values of \(a\), including 1). In fact, starting with the more complicated trinomials where \(a\neq1\) is preferred, as it is easier to then explain how work can be reduced and omitted when \(a=1\), rather than trying to build up to more steps (which ultimately feels like they are two different methods, rather than one method with a simpler/shorter version).

Disclaimer

Please do not misinterpret this article as a suggestion that all other methods of factoring are inadequate and should not be revealed to students. In fact, students should have the ability to learn about and use multiple methods, so they can determine what is the best of their individual use.