Factoring quadratics and linear equations

Factoring a quadratic equation involves finding two linear equations whose product is the quadratic equation. This is an example where mathematics teachers often act as if (a) there is one method of solving and (b) there is one solution.

The AC Method

The “one method of solving” strategy usually goes something like this: If the \(x^2\)-coefficient is 1, then find two numbers whose sum is the \(x\)-coefficient and whose product is the constant. If the \(x^2\)-coefficient is something other than 1, then use the AC method, which is best explained with an example.

For instance, let’s say we want to factor \(y = x^2 + 6x + 9\). 9 has factors of 3 and 3, which add to 6, so we’re done: \(x^2 + 6x + 9 = (x + 3)(x + 3) \). Let’s say we want to factor \(y = x^2 + 7x + 12\). 12 has factor sets (1, 12), (2, 6), and (3, 4), the last of which have the sum of 7. Hence, \(x^2 + 7x + 12 = (x + 3)(x + 4)\).

The AC method is based on a similar strategy, with an additional step. Actually, the AC method is the same strategy; the method above is just a simplification based on the fact that the coefficient on the first element is 1. The AC Method:

  1. Multiply the first and third coefficients (that is, \(a\) and \(c\)).
  2. Find two factors of the result that add to the second coefficient.
  3. One of those two factors will have \(a\) as a factor. Divide by \(a\) and pair it with \(x\). This is one of the factors.
  4. Pair the other factor with \(ax\). This is the other factor.

For instance, let’s say we have \( 2x^2 – x – 6 \).

  1. \(2 \cdot -6 = -12\)
  2. -12 has factors of -4 and 3, which add to -1.
  3. -4 has 2 as a factor. \(\frac{-4}{2} = -2\), so \(x – 2\) is a factor.
  4. \(2x + 3\) is a factor.

Hence, \( 2x^2 – x – 6 = (x-2)(2x + 3) \). Again, note that the first method is the AC method in the case that \(a = 1\), so much of the time in mathematics class, we’re teaching two “methods” and letting students work out that it’s really just the same thing.

All this aside, a weakness of the AC method is that it relies on there being factors that satisfy the criteria, which in turn relies on artificially constructed problems. Another is that implies that there’s a single correct factorization. There is often a “best” factorization, but there are always others; we can get other factorizations simply by multiplying the first factor and dividing the second factor by the same amount. For instance, in the example above, \( 2x^2 – x – 6 = (2x-4)(x + \frac{3}{2}) = (\frac{x}{2} – 1)(4x + 6) \).

The Quadratic Formula

The Quadratic Formula will yields factors. It has the disadvantage that it “looks scary”; it’s very mechanical. It has the advantage that it will allow us to factor any quadratic, not merely those with simple (particularly integer) factors.

The Quadratic Formula: \[ x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \]

In our example above, we solve for \(x\) as follows: \[x = \frac{1 \pm \sqrt{(-1)^2 – 4(2)(-6)}}{2(2)} \\ = \frac{1 \pm \sqrt{1 – (-48)}}{4} \\ = \frac{1 \pm 7}{4}\]

This yields the solutions \(x = 2\) and \(x = \frac{-6}{4} = \frac{-3}{2} \).

Now, this returns solutions (that is, the points at which \(y = 0\)) rather than factors, which was the stated goal of the problem. However, it should be easy to see the relationship between \(y = 2x + 3\) and \(x =  \frac{-3}{2}\). This leads us to a generalization: The solutions of a quadratic equation can be written in the form \(x = \frac{-b}{m} \), where \(b\) and \(m\) correspond to the coefficients in the slope intercept form of linear equations, that is, \(y = mx + b\). Our solutions above yield \(y = x – 2\) and \(y = 2x + 3\), which are the same factors we got from the AC method.

There is the additional wrinkle that this method works best when the solutions for \(x\) are simplified as much as possible; as we saw above, \(y = 4x + 6\) is also a factor of \(y = 2x^2 – x – 6\), but \(y = (x-2)(4x+6) = 4x^2 -2x-12\), which is twice our target.

However, what this method reinforces is the connection between quadratic factors and linear equations, since the factors of a quadratic equation are, in fact, linear equations. This might not normally be noticed by students, particularly when the AC method is taught as “the” way to get “the” answer.

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