Today we learned about factoring quadratic trinomials and factoring difference of squares.

Factoring Difference of Squares

Say we have \(ax^2 - b\).

Factoring is the opposite of distributing, so we just need to find something that distributes to \(ax^2 - b\).

Usually, you want to factor out a GCF before you do anything else. Once you do that, we get something like: \(c(\frac{a}{c}x^2 - \frac{b}{c})\).

Then, you know that the two coefficients are squares of some other numbers, so you just factor it as: \(c(\sqrt{\frac{a}{c}} + \sqrt{\frac{b}{c}})(\sqrt{\frac{a}{c}} - \sqrt{\frac{b}{c}})\) (here, you could technically take the \(\sqrt{\frac{1}{c}}\) out and cancel it with the c in front, but that’s not fully simplified.)

This process goes basically like this: I want to factor a difference of squares. When you distribute a difference of squares, it squares the coefficients, so if I want to factor it, or go backward, I need to square root the coefficients.

Factoring a quadratic trinomial

Say your quadratic trinomial looks like this: \(ax^2 + bx + c\).

You want to factor it. First, let’s look at the case where \(a = 1\), because it’s simpler and probably a better place to start.

Factoring when \(a = 1\)

Factoring is the opposite of distributing, so let’s look at what happens when we distribute.

When we distribute \((x + r)(x + s)\), we get \(x^2 + x(r + s) + rs\). We notice that \(b = r + s\) and \(c = rs\). Therefore, we should find two numbers, r and s, such that \(r + s = b\) and \(rs = c\).

There’s a few tricks that can help here:

First, look at whether b or c is positive or negative. If c is negative, one of \(r\) or \(s\) will be negative. That means that \(b\) will be the difference of two positive numbers. If c is positive, \(r\) and \(s\) will both be either positive or negative. The way to tell if they’re both positive or negative is to look at \(b\).

After that, look at \(c\). If \(c\) has a lot of divisors, then your best bet is to look at \(b\). If \(b\) is a weirder number to get, you can try looking for numbers that add up to \(b\). If \(c\) is more on the prime side (or better yet, prime), you can just test out all of the divisors of \(c\).

Basically, look for the things that are unique, and then use that uniqueness to find the right numbers.

Factoring When \(a \neq 1\)

We only learned how to do it if you could do the first option below and have all integers (basically same thing as factoring out a GCF).

More generally, you have 2 options:

First, you can factor an \(a\) out of the whole thing and risk having fractions.

Second, you can factor it while taking into account that \(a\) isn’t \(1\). Basically, you will factor it into \((dx + e)(fx + g)\) instead.

Then, \(df = a\) and still, \(c = eg\), but \(b\) has changed. \(b\) is now \(dg + ef\).

When I solve this type of quadratic, I usually vary \(d\) and \(e\) (dependent on \(a\)) and try out a lot of combinations for \(f\) and \(g\). If it makes more sense to vary \(f\) and \(g\) instead, it’s possible to do that too.

Basically every trick mentioned in the above section still works. Just watch out for the weird \(b\) and you’ll be fine.

Conclusion

This is probably the last factoring thing we’re going to do. We might do some stuff about solving quadratics, but I can’t gaurantee that this will be in the same unit.