Does anybody know how to do algebra II, and some ways to help know/remember it?

I'm pretty sure it's what it's called.

Factoring Polynomials and stuff like that.

alexanyways : I've taken a lot of math courses over the last few years, so I might be able to help some if you can ask any specific questions.

soxfan849 : I just need to know the basics.

I've been out of Algebra 2 for like 4 years, so I don't really remember enough off hand to give you any real help. If you have specific problems that you need someone to show you how to do, I could help with that. Otherwise I'm pretty much useless to you.

I don't know what else to say, except that you could try using google to see if there are any good websites to explain the subject matter. This was the first site I found with a quick google search, and it looks like it might be somewhat helpful.

Algebra II is a pretty general subject matter. Could you narrow it down to something like a specific lesson?

Someone70 : Factoring Polynomials in peticular is what I need help with.

Okay.

Can you give me an example problem that you have trouble with? I'll explain and work it out best I can.

Or are you confused with just the concept of factoring in general and/or why we have it?

Someone70 : I'm confused with the whole thing, I don't have much problem with the concept of it, or how it will be useful in the future, but the important thing is that we have it and I don't understand a lick of anything about it.

(sorry this is coming a bit late, but I was out for most of today. x3 )

Well, factoring is just reverse distributive property. That's pretty much why it's there. If you look at:

2(x + 3) => 2x + 6

Then,

2x + 6 => 2(x + 3)

is just the opposite of that.

So factoring polynomials is just that, but with a few more steps. There are actually several methods you can use, but I'm guessing that you need help with all of them? I'll go from easiest to hardest.

1. Greatest Common Factor

This is the way that you'll recognize factoring the easiest. Simply put, if every term can be divided by a certain number or variable, then you can factor it out. Take a look at this:

6x^2 + 4x + 2

See how you can divide each of the coefficients by 2? If you do that, you can bring out the 2 and you get:

2(3x^2 + 2 + 1)

Not too hard is it? You can do the same with variables, too.

y^5 + 4y^4 + y^2 => y^2(y^3 + 4y^2 + 1)

Don't forget that when you take y^2 out of that last term, you're left with 1.

2. Difference of Squares

Now, difference of squares is a special case that only works with binomials whose first and second terms are both perfect squares and they are being subtracted. If you look at

4x^2 - 9

you can see that the first term is 2x squared and the second term is 3 squared. When you see this, then you instantly know you can put it in this formula:

(a + b)(a - b)

where a is the positive square root of the first term and b is the positive square root of the second term. So, using the example above, the way to factor it would be like this:

(2x + 3)(2x - 3)

If you were to FOIL that, you would see that the middle terms cancel out and you're back to the question. If you see that one of the resulting binomials is also a difference of squares, be sure to continue factoring that, too.

3. Sum/Difference of Cubes

This is yet another special case that only works with binomials whose first and second terms are both perfect cubes. It doesn't matter if they're being added or subtracted. Take a look at this:

y^3 + 27

The first terms is y cubed and the second term is 3 cubed. In this situation, you use the formula:

(a +/- b)(a^2 -/+ ab + b^2)

So whatever the sign is in the problem, that sign will be used in the first binomial. Then, for the first sign in the trinomial, you switch it. The last sign is always positive. Factoring the problem above, you get:

(y + 3)(y^2 - 3y + 9)

This formula is a little more complicated than difference of squares, so remember it well.

4. Quadratic Trinomials (The "Bottoms-Up" Method)

First, before I start, remember that the general form of a quadratic equation is:

ax^2 + bx + c

where a, b, and c are all coefficients.

Let's say you have the following:

2x^2 - 7x + 5

You can't use any of the above strategies for this, right? So what you do here is something that I like to call "bottoms-up" method. You'll see why, later.

The first thing you do is multiply a and c (in this case, you get 10). Now you have to ask yourself, "What two numbers can I multiply to get 10 and add to -7?" The answer happens to be -2 and -5. Now "divide" those two numbers by the "a" term and you should get the fractions -1/1 and -5/2. Then, just read those fractions from the "bottom-up" and put them into your binomial factors. So your result would be:

(x - 1)(2x -5)

And magically enough, that's the answer! So why did you do that? Well, foil that out and you'll see that you ended up right where you started! See? Magic math.

5. Completing the Square

This isn't a method that I would normally recommend, unless you need an equation in vertex form or your teacher asks specifically for it.

But, let's say you have the polynomial

x^2 + 6x - 2

You can't really use the "bottoms-up" method since nothing multiplies to -2 and adds to 6, right? So you could try completing the square. This method is used when you want to get it in the form:

a(x - h)^2 + k

where (h,k) is the vertex of your parabola. This is also known as vertex form. It seems like a complicated form, but it's actually not too difficult to get to. How you're going to do it is by first taking b (6 in this case) and dividing it by 2. Then you take that number (3) and square it (9). You then add that too your equation (it helps to stick it in between the 6x and -2). But don't forget to keep things equal. To offset the adding of 9, you should subtract 9 from your equation. So now you get:

(x^2 + 6x + 9) -2 -9

(parenthesis added for emphasis) To turn it into a squared binomial is simple. Take the original b/2 number (3 in this case) and use that for h in the vertex form. So it would look like this:

(x + 3)^2 - 11

There, that's pretty much all you need to know about factoring. If you have any more questions, please feel free to ask.