>> I want to look at a couple examples

of factoring using our general factoring techniques.

Here's an overall strategy that we can look

at when we factor polynomials.

Let's go through these steps one at a time here.

The first one, always look for a greatest common factor.

If there's any common factor other than 1,

it needs to be factored out to the front of the polynomial.

How many terms are in the polynomial?

If there's 4 terms, you're probably going to use grouping.

If it's 3 terms, I prefer the trial and error method.

We're talking about trinomials now.

So trial and error is the backwards FOIL.

And of course, if you prefer one of the other methods,

the most common one would be when you split

up the middle term and using grouping.

But any other method would be perfectly fine there

if you have a favorite one.

And if it's not 4 terms or 3 terms, of course,

it could be greater than 4, but these are the most common ones.

There'll be at least 2 terms.

So this is a possibility.

And if there's 2 terms, it's probably going to be one

of these cases, difference of squares,

which we know has a nice, simple pattern, a plus b,

a minus b for the 2 factors.

Sum of the squares never factors.

The only possibility would be a GCF.

And that, of course, should be factored out in the first step.

There's 2 other possibilities, the difference

of cubes and sum of cubes.

And each of those has its own special formula.

They're very closely related.

There's a very -- a pattern to each of these that you'd look

at if you're memorizing those.

There's really no way around these.

You can't really use trial and error for the sums

and differences of cubed like you could

for the difference of squares.

All right.

And then when you think you're done factoring,

look at all your factors and see if any

of them could be factored further.

You know, maybe one of the factors is another difference

of squares that you can factor.

Maybe there's a GCF that you've missed in the first step

that can be factored out of one of the terms or one

of the binomial factors.

And lastly, you can check your answer

by multiplying your factors back together to see

if you get the original polynomial.

That will not net you if you've factored it completely,

you know, if you missed anything from step 3.

But it will at least tell you if the factors that you have

so far are correct or if you're on the right track.

All right.

Let's try this out.

It's not as bad as it sounds.

Factor completely, 6x cubed minus 27x squared minus 15x.

My first step is to look for a greatest common factor.

So I'm asking myself is there any greatest common factor

or any common factor other than 1 that I can factor out front?

Look at the variable and the coefficients separately.

When I look at the variable,

I notice the x shows up in all 3 terms.

So that means x is going to be part of my GCF.

When you look at the exponents of each of the terms,

the variables on each of the terms, you're going

to want the smallest of all 3.

So x to the 1 is the only power of x I can factor out.

The coefficients, 6, 27, and 15, all have a factor of 3.

So my GCF, the answer is yes, there is a GCF other than 1

that I have to factor out front, so 3x.

And when you're factoring this out,

you can go straight to this step.

You put the 3x in front of the parentheses.

And there's going to be as many terms inside the parentheses

as you have in your original polynomial.

And you simply divide 6x cubed by 3x to get the first position,

similarly for the second and the third.

If you're new to this, you might find it helpful to write each

of these original terms as a product

of the GCF and whatever's left.

I didn't leave myself much room there to squeeze that in.

But 3 times 2 would give me 6,

and x times x squared would give me x cubed,

3x times 2x squared would give me 6x cubed.

That 2x squared, that's exactly what you'd get

if you divide 6x cubed by 3x.

For some reason, I always -- in my own head, I always think

of it as multiplication rather than division.

What do I need to multiply this 3x by to get the term?

In this case, I would need to multiply by 9 to get 27

and x times x to get x squared.

So 27x squared divided by 3x is 9x.

And that would go in the middle.

And of course, I didn't factor out a negative.

About the only time I would factor out a negative is

if it was on the first term.

So that subtraction stays, same with this one.

And the 15x divided by 3x is going to be 5.

So a 5 would go in here.

So to complete my picture above, I would look at my 3x,

and I would think I have to multiply it by 5 to get 15x.

All right?

So you can clearly see from the factors

above when I distribute the 3x, I'm going to get each of these.

And it turns back in to the original polynomial.

So factoring out a greatest common factor, we're just kind

of looking at the distribution property backwards.

And now, because I have 3 terms, I'm looking at this.

It's a product of 2 binomials using my trial and error method.

Don't forget about the GCF that you pull out front.

That has to come along at every step,

or at least make sure you put it in your final answer.

Trial error, reverse FOIL, the product of the first terms,

this one and this one, have to give me 2x squared.

And that one, because 2 only has factors 2 and 1,

and we split the x squared up into x and x, we get 2x and 1x.

So there's only 1 possibility here.

I might as well go ahead and put that in.

That takes cares of my first -- my product of my first terms.

The product of the last terms has to give me a negative 5.

Don't forget about the sign.

So I'm going to have -- that's supposed to be a comma --

1 and a 5 with opposite signs, 1 positive and 1 negative,

to get a negative 5 when I multiply them.

I'm not going to worry right now about which one's going

to be positive and which one's going to be negative.

I'm going to do my trial and error here.

Let me try, for example, the --

putting the 5 first and the 1 second.

The order's going to matter since the 2x

and the x are different and the 5 and the 1 are different.

It's going to matter what order we put these in.

So, that'll be part of our trial and error.

I've already verified that these products, the first

and last products, will give me the first and last terms here.

The only ones left are the outer and the inner, the O and the I.

The outer product is out here, and the inner product is here.

So in this case, I would have 5 times x for my inner product

and 2x times 1, which is 2x, for my outer.

They're going to have opposite signs, 1 positive, 1 negative.

So I can either get a positive 3x or a negative 3x,

depending on which one is positive, which one is negative.

That's not going to work.

I need a negative 9x.

So --

Cross that out.

I don't have any other ones to choose from.

I just have to try switching the order.

Now my inner product is going to be 1 times x,

which is just x. My outer product is going

to be 2x times 5, which is 10x.

That will have a difference, if 1 is positive and 1 is negative.

I will have a difference of 9x, so either positive 9x

or negative 9x, depending on where I put the positive

and where I put the negative.

I want negative 9x.

So the 10x has to be negative, and the 1x has

to be positive to give me that.

And I will get that by putting a negative on the 5

and a positive on the 1.

So that looks like it's going to work.

My factors have -- well, 3x, if I consider that 1 factor,

2x plus 1 is a factor, and x minus 5 is another factor.

And that's as far as I can go.

Let's look at 1 last example.

Factor completely 16x to the 4th minus y to the 4th.

My first question, just like last time,

is there a greatest common factor that I can factor out?

Well, the variables are different.

So, there's no common variable factor.

And the coefficients are 16 and 1 or negative 1.

So my greatest common factor is 1.

So it's not that there's none.

But since it's 1, there's nothing to factor out.

So I'm going to just go ahead and say no.

So nothing to do on step 1, other than check.

Step 2, well, there's 2 terms.

So I'm going to look

for possibly a difference of squares.

Definitely not a sum of squares, because of the minus.

This looks like a good candidate for difference of squares

because I have perfect squares for my coefficients, and I have,

as we'll see later,

what's important here is the exponents are multiples of 2.

They're even numbers.

So if I can show that this is something squared minus

something squared, it's going to factor

as the a plus b, a minus b. All right.

So what squared minus what squared do I have here?

The coefficient, 16, is 4 squared.

So I put a 4 inside the parentheses.

And the variable, well, x by itself won't work

because this would just be x squared.

But remember your rules of exponent, 2 really actually,

if I split the exponent up among the 4

and the variable here first.

And then if I put an exponent on the variable,

I'm going to have to multiply these.

So I need 2 times 2 to give me 4.

So my a is 4x squared everything inside the parentheses here.

This one's a little easier.

There's no -- the coefficient is just 1.

Again multiplying these exponents, y squared,

squared is y to the 4th.

So I can write this as a squared minus b squared.

And so, this is going to factor

as a plus b times a minus b. Here's my a plus b times a minus

b. That's supposed to be a square there, y squared.

Now these are a little more complicated

than the factors I had in the previous example.

This one is a sum of squares.

These are 2 perfect squares.

So that one cannot be factored further.

This one is a difference of squares.

So I have another difference of squares.

I'm going to have a different a and b now.

Now I need to square a 2 to get 4 and x to get x squared.

And this is just y squared.

So my a is 2x.

My b is y. And plugging back into this formula one more time,

a plus b, a plus b, a minus b, and don't forget

about the 4x squared plus y squared from the previous step.

That was not changed.

That's still part of our answer.

So now nothing can be factored any further.

We are done.

I will leave it to you if you want

to multiply these back together and then multiply by this one

to see if you get the original polynomial back again.