In this class, they (not including me, I wasn’t there) learned about factoring. This most likely isn’t the factoring you’re thinking about.

We learned three things: factoring a GCF out of a list of terms, factoring a GCF out of polynomials, and factoring 4-termed polynomials.

Factoring a GCF (greatest common factor) Out of a List of Terms

Just in case somebody doesn’t know, a list of terms looks like this:

\[2xy^5, 12x^2y, 500xyz^3, 20xyz\]

Hopefully, you know how to find a GCF already.

If you don’t, here’s a quick explanation:

Step 1:

Prime factorize all of the numbers.

Step 2:

List out all of the different primes from all of the numbers.

Step 3:

Find the highest exponent attached to each of the primes. Then, attach that exponent to it’s corresponding prime.

Step 4:

Multiply everything together, and you have the GCF.

Generally, if you find yourself with a whole lot of variables or irreducible constants, pretend those are regular numbers and just do the same steps.

This works here too. When you prime factorize all of the terms, treat the variables (and if you see them, irreducible constants) as regular primes.

Factoring a GCF Out of a Polynomial

You can probably figure this one out :)

Solution Treat the polynomial as a list. Then, do the GCF thing with the list, and then, combine the list back together into a polynomial.

Factoring a Cubic Quadrinomial

This method only works sometimes.

Basically, you treat the \(x^3\) and \(x^2\) terms as one polynomial, and the \(x\) and constant terms as another. Then, you can use the GCF thing on both polynomials.

If you do this, you are guaranteed to have \(ax^2(bx + c) + d(ex + f)\).

IF \(b=e\) and \(c=f\), you can combine those again by factoring out \((bx + c)\) from both of them.

This doesn’t only work if you group it this way.

Generalization

Generally, you can split things up in any way and then factor them as long as the number of groups multiplies together to the number of terms and the terms mesh together perfectly.

For example, say we have a six-termed polynomial:

\[ax^5 + bx^4 + cx^3 + dx^2 + ex + f\]

The number of terms is \(6\), so we can split it up into \(2\) and \(3\), because \(2 \cdot 3 = 6\).

Then, it would look something like this:

\[x^4(ax + b) + x^2(cx + d) + x^3(ex + f)\]

and then, if the terms mesh together,

\[(gx^4 + hx^2 + jx^3)(kx + l)\]

(if you noticed I didn’t use \(i\) as a coefficient, good job. \(i\) is actually \(\sqrt{-1}\), so I couldn’t use it as a variable.)

We could also factor it into more layers than just 2.

If we used a eight-termed polynomial instead, we could have three layers. First, we would seperate the polynomial into 4 binomials. Then, we could combine those binomials into 2 groups of 4, where each group of 4 has 2 layers. Finally, we could combine those last 2 factored polynomials into something with 3 layers.

This doesn’t stop here: you can make the number as big as as you want, and factor it into as many layers as you want.


As the title suggests, this isn’t all of factoring. There is still a lot to do, like factoring quadratics (in different ways than mentioned here) and probably completing the square and the quadratic formula.