Factoring Quadratic Equations the Easy Way

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Factoring Quadratic Equations the Easy Way

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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

    \[ax^2 + bx + c\]

that can be factored into integers, we can find a factorization of the form

    \[(x + \frac{c}{f_1})(x + \frac{f_2}{a})\]

such that

    \[a\cdot c = f_1 \cdot f_2\]

and

    \[f_1 + f_2 = b.\]

It follows that, given a trinomial/quadratic equation of the form

    \[ax^2 + bx + c = 0\]

that can be factored into integers, we can find roots of the form

    \[x_1 = -\frac{c}{f_1}\]

and

    \[x_2 = \frac{-f_2}{a}\]

such that

    \[a\cdot c = f_1 \cdot f_2\]

and

    \[f_1 + f_2 = b.\]

 

Method to Teach to Students 

Quadratic Expressions: You are given a quadratic expression ax^2 + bx + c. Our process for finding factors is as follows.

1. First, multiply a and c. List all of the factor pairs, positive and negative, that are equivalent to a\cdot c.

2. From the list of factor pairs you determined, find the set that, when the elements are added together, equal b. Call this pair (f_1 ,f_2). This gives us that

    \[a\cdot c = f_1 \cdot f_2\]

and

    \[f_1 + f_2 = b.\]

3. Once you determine f_1 and f_2, rewrite your quadratic expression with f_1 and f_2 substituted for b:

    \[ax^2 + bx + c\]

    \[\implies ax^2 + (f_1 + f_2)x + c\]

4. Now group the expression so that f_1 is couples with ax^2 and f_2 is coupled with c. It is important to note that f_1 and a must share a factor, just as f_2 and c must share a factor.

    \[ax^2 + (f_1 + f_2)x + c\]

    \[\implies (ax^2 + f_1 x) + (f_2 x + c)\]

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 f_1 as \frac{ac}{f_2}, which comes from

    \[a \cdot c = f_1 \cdot f_2\]

Once we do this, we have

    \[(ax^2 + f_1 x) + (f_2 x + c)\]

    \[\implies (ax^2 + \frac{acx}{f_2}) + (f_2 x + c)\]

    \[\implies ax {\bf(x + \frac{c}{f_2})} + f_2 {\bf (x + \frac{c}{f_2})}\]

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:

    \[ax(x + \frac{c}{f_2}) + f_2 (x + \frac{c}{f_2})\]

    \[\implies (x + \frac{c}{f_2})(ax + f_2)\]

To isolate x in the second set of parentheses, we factor out the a and get

    \[(x + \frac{c}{f_2})(ax + f_2)\]

    \[\implies a(x + \frac{c}{f_2})(x + \frac{f_2}{a})\]

We can see that our factors are (x + \frac{c}{f_2}) and (x + \frac{f_2}{a}).

 

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.

    \[a(x + \frac{c}{f_2})(x + \frac{f_2}{a}) = 0\]

    \[\implies (x + \frac{c}{f_2}) = 0\]

 and 

    \[(x + \frac{f_2}{a}) = 0\]

Then

    \[x_1 = -\frac{c}{f_2}\]

and

    \[x_2 = -\frac{f_2}{a}\]

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 ax^2 + bx + c = 0, that can be factored into integers, such that a \cdot c = f_1 \cdot f_2 and b = f_1 + f_2, we have that

    \[ax^2 + bx + c = 0\]

    \[ax^2 + (f_1 + f_2)x + c = 0\]

    \[(ax^2 + f_1 x) + (f_2 x + c) = 0\]

    \[(ax^2 + \frac{acx}{f_2}) + (f_2 x + c) = 0\]

    \[ax(x + \frac{c}{f_2}) + f_2 (x + \frac{c}{f_2}) = 0\]

    \[(x + \frac{c}{f_2})(ax + f_2) = 0\]

    \[a(x + \frac{c}{f_2})(x + \frac{f_2}{a}) = 0,\]

thus

    \[x_1 = -\frac{c}{f_2}\]

and

    \[x_2 = -\frac{f_2}{a}.\]

 

Note: Another interesting investigation would be to see if this works for quadratic expressions that are not factorable into integers.

 


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