How do I factor quadratics when the coefficient of x^2 isn't 1?

  • MOD

    Module 4 Week 3 Day 12 Your Turn Explanation

    Some students have been asking about factoring quadratics when the leading coefficient isn't \(1\) (such as, \(2x^2 - 7x + 6\)), so I thought it would be helpful to share the method for doing so. It's definitely harder than factoring monic quadratics (that look like \(x^2- 7x+6\)), but fortunately there are still some cool ways to figure out how to factor.

    If I have a quadratic in which the coefficient of the \(x^2\) term is \(2\), like \(2x^2 - 7x + 6 = 0\), a good way to figure out how to factor it (...if there even is a clean way to factor it...) is to think about how the factorization might look like. Before, the coefficient was just 1, and factoring a quadratic like \(x^2 + 3x - 4\) was easier because we knew the factorization would look something like \((x-a)(x-b)\), and so we just need to figure out the numbers \(a\) and \(b\) so that \(a+b = -3\) and \(ab = -4\).

    So what would the factorization of \(2x^2 - 7x + 6\) look like? It probably wouldn't look like \((x-a)(x-b)\) anymore, because then when you expand that, the first term is \(x^2\), not \(2x^2\). How should we get the \(2\) in there? Perhaps instead, the factorization looks like \((2x-a)(x-b)\). That would work!

    The problem now is, how do we find the numbers \(a\) and \(b\) that make this work? In other words, we need to find \(a\) and \(b\) so that:

    $$(2x-a)(x-b) = 2x^2 - 7x + 6$$

    If we think about how we would expand \((2x-a)(x-b)\), you can see that we still have \(ab = 6\), so that's easy. But would we still have \(a+b = 7\)? You can probably see that it's not so simple now, because of that extra \(2\)! Do you see that we actually have \(a + 2b = 7\)? So instead of \(a+b = 7\), we need to get \(a+2b = 7\)...

    If that's not clear why, let's expand \((2x-a)(x-b)\). You can use the distributive property or even "FOIL". You should get that \((2x-a)(x-b) = 2x \cdot x - 2x\cdot b - a \cdot x + a \cdot b\), which is just \(2x^2 - (a+2b)x + ab\).

    Aha! That's why we still have \(ab = 6\), but now we need \(a+2b = 7\). To recap:

    • We need to find two numbers that multiply to \(6\) (the constant term), AND
    • If you multiply one of these numbers by the coefficient in front of the \(x^2\) (the "leading coefficient"), and add it to the other number, we should get \(7\).

    It might take some trial and error, but eventually you might come up with \(2\) and \(3\). Their product is definitely \(6\), and when you double \(2\) and add it to \(3\), you get \(7\).

    Now we need to put the \(2\) and \(3\) back into the \((2x-a)(x-b)\) form. But which number is which, does the order matter? Remember, to get the \(7\), we had to double the \(2\), not the \(3\). The number that gets multiplied by the \(2x\) is \(b\), so we need to put \(2\) into \(b\) and \(3\) into \(a\), to get \((2x-3)(x-2)\). And that's how we can get the factorization!

    You can try practicing this on these quadratics:

    1. \(3x^2 - 13x + 12 = 0 \) (Hint: You need to find two numbers that multiply to \(12\), and so that when you multiply one of them by \(3\), they add to \(13\).)
    2. \(5x^2 + 29x - 6 = 0\)

    As a final comment, if it seems really hard to factor the quadratic nicely, don't lose hope! You can always use the quadratic formula! As an example, let's say I had a lot of trouble factoring that original quadratic: \(2x^2 - 7x + 6 = 0\). The quadratic formula works for ALL quadratics! It says that if you have a quadratic \(ax^2 + bx + c = 0\), then you can solve for \(x\) as:

    $$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

    For the quadratic \(2x^2 - 7x + 6\), we see that \(a = 2\), \(b = -7\), and \(c = 6\). So we can just plug it in to the formula:

    $$\begin{aligned} x &= \frac{7 \pm \sqrt{(-7)^2-4(2)(6)}}{2(2)} \\ &= \frac{7 \pm \sqrt{49 - 48}}{4} = \frac{7 \pm 1}{4} \\ &= \frac{7 + 1}{4}\ \text{or}\ \frac{7-1}{4} \\ &= 2\ \text{or}\ \frac32 \\ \end{aligned} $$

    Notice that these are the same answers you get if you solve the factored version, \((2x-3)(x-2) = 0\). That's cool! In fact, you can actually use the quadratic formula to figure out the factorization once you figure out the values of \(x\) that work. Since we figured out using the formula that \(x = 3/2\) or \(x = 2\), we can come up with a simple factorization with these solutions:

    $$\left(x - \frac32\right)(x-2) = 0$$

    But hey, if we multiply each side by \(2\), we get the original factorization we came up with!

    $$(2x-3)(x-2) = 0$$
    That's pretty cool! To recap:

    • If you need to factor a quadratic \(ax^2 + bx + c\) where \(a\) isn't \(1\), try to find two numbers that multiply to \(c\), and such that their sum (with one of them multiplied by \(a\)) is \(-b\).
    • If it's not working, just use the quadratic formula! This is a surefire way to get the solutions to the quadratic.
    • If you still want to figure out how to factor it, the solutions you get from the quadratic formula can tell you how you could have factored the quadratic!

    Hopefully this helps with everyone's quadratic solving endeavors. Happy learning!