When I was in high school, I remember having acute anxiety about having to factor quadratic equations on tests and quizzes. This was the ambiguous section of high school algebra when all trinomials were magically factorable into nice, whole integers – the section before learning the ever-useful Quadratic Formula.
The reason I had this anxiety was that it seemed as though there was no precise algorithm for factoring trinomials. You simply had to play with your a, b, and c’s until you found two numbers that multiplied to equal c, and added to equal b. Easy, right? But the problem is that a lack of precision is stressful at that age. So how do you cope?
One of my student’s teachers showed her class how to algorithmically find the factors of a integer factorable trinomial. I’ve formalized it as a theorem for specific trinomials, and I’d love to share it with you today. I’ve also provided the theorem’s quick proof, and perhaps most importantly, the general method for finding factors based on this theorem. This nice little theorem should find its home as a staple in algebra textbooks.
Integer Factorable Quadratic Theorem
Given a trinomial/quadratic expression of the form
that can be factored into integers, we can find a factorization of the form
It follows that, given a trinomial/quadratic equation of the form
that can be factored into integers, we can find roots of the form
Method to Teach to Students
Quadratic Expressions: You are given a quadratic expression . Our process for finding factors is as follows.
1. First, multiply and . List all of the factor pairs, positive and negative, that are equivalent to .
2. From the list of factor pairs you determined, find the set that, when the elements are added together, equal . Call this pair . This gives us that
3. Once you determine and , rewrite your quadratic expression with and substituted for :
4. Now group the expression so that is couples with and is coupled with . It is important to note that and must share a factor, just as and must share a factor.
5. Pull out a common factor from the elements of the first set of parentheses.
6. Pull out a common factor from the elements of the second set of parentheses. Now the elements in the first and second set of parentheses should be the same.
We can see this by re-writing as , which comes from
Once we do this, we have
Now we have that the quantities in both sets of parentheses are equal. We can now factor out the expression inside the parentheses and get our factors:
To isolate in the second set of parentheses, we factor out the and get
We can see that our factors are and .
Quadratic Equations: Finding roots of a quadratic equation follows the same process, but we take it one step further; we set our factors equal to zero and solve for the two possible solutions.
are the roots of the quadratic equation.
Quick Theorem Proof
This actually follows the same algorithm as the Method to Teach to Students. I’ll show it here for quadratic equations.
Given a quadratic equation , that can be factored into integers, such that and , we have that
Note: Another interesting investigation would be to see if this works for quadratic expressions that are not factorable into integers.