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# Factoring A Trinomial (Illegal Method)

Factoring trinomials.

If I didn't lose you at that, then you're here for help, which is what I shall give (hopefully).

Factoring trinomials is easy enough, as long as, in an Ax+Bx+C trinomial, Ax=x^{2}. But factoring out something like 4x^{2}+29x+30 can wear on your brain. And when you have thirty of those questions in a row, and you need to get them done quickly so you can get to other homework or activities, then factoring it out mentally is a real pain. We're all in luck, though, as there is a method for factoring trinomials that can eliminate all of that nasty guessing and checking.

First of all, the leading coefficient, your A, cannot be 1. Also, if A is a negative, then you'll need to factor it out, but we'll get to that. We'll need an example trinomial first.

3x^{2}-2x-5

Okay, not too hard. You could probably do this one in your head, but who has time for that, right? Thinking is overrated anyway. So let's make it easy.

**Step 1:** Multiply the leading coefficient, A, by C, and form a new trinomial. Like so:

x^{2}-2x-15

Woah. What just happened there? You took one number from a problem and multiplied it by another number* in the same problem*? Not on the other side of the equals sign or anything? Yeah. It's okay, I was skeptical too at first. We'll be undoing that later anyway. Besides, now we're back to our standard x^{2}, so factoring this will be cake.

**Step 2:** Factor your new trinomial.

(x-5)(x+3)

We're not done yet, though. Remember that crazy move we did earlier, you know, the one that broke mathematical law? Well, we'll undo that here.

**Step 3:** Undo the "illegal move" by dividing your coefficients by your old leading coefficient (A).

(x-5/3)(x+3/3)

Wait, no, stop! Fractions in the binomials? What are you going to do now? Easy. Relax about the fractions, and remember that you are only dividing. Reduce your fractions (if you can).

**Step 4:** Reduce your fractions.

(x-5/3)(x+1)

Now it's a little better, but you still have that pesky 5/3. We can't have that. But the good new is we can get rid of it.

**Step 5:** Clear any fractions by moving the denominator in front of the x.

(3x-5)(x+1)

Wait... if you solve that... just a little FOILing here and there and... no... Wow. There, above this, are your factors. To check, just multiply that, and you get:

3x^{2}-2x-5

Tada. Just a little finagling and you can get the factors of your trinomial without hurting your head. Oh, and if you're worried about that "factoring out the negative" thing I talked about earlier, don't be. Here's an example of how to factor a trinomial with a negative leading coefficient, using the illegal method.

-6b^{2}-25b-25

-1(6b^{2}+25b+25)

-1(b^{2}+25b+150)

-1(b+15)(b+10)

Bam.

This will work for all basic trinomials (meaning I haven't seen one that this doesn't work for, but I bet they're out there. Math is sneaky like that). Use it for yourself, tell your friends, impress your teachers, or just continue factoring in your head (I'll admit it, I do).

## Comments

I was skeptical of it, too!

What you're doing is:

* multiplying the whole thing by a, so you have (1/a) [a^2 x^2 + ab x + ac]

* replacing ax with X: (1/a)[X^2 + bX + ac]

* factorising: (1/a)(X - stuff)(X - stuff)

* returning to x: (1/a)(ax - stuff)(ax - stuff)

* and dividing the a out in the appropriate place.