1) x^2 - x - a - a^2

2) x^4 - 3x^2 - 4

3) x^4 - 8x^2 + 16

4) xy + 2 - 2x - y

5) 5s^2 - 16s + 12

6) 6a^3 - 7a^2b - 2ab^2

How do I factor these completely? I'm really lost right now

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- Oct 21st 2007, 03:21 PMDunit0001Factoring?
1) x^2 - x - a - a^2

2) x^4 - 3x^2 - 4

3) x^4 - 8x^2 + 16

4) xy + 2 - 2x - y

5) 5s^2 - 16s + 12

6) 6a^3 - 7a^2b - 2ab^2

How do I factor these completely? I'm really lost right now - Oct 21st 2007, 03:33 PMgalactus
Practice is what factoring requires.

Quote:

How about -4 and -4

Notice the difference of two squares?

- Oct 21st 2007, 03:47 PMtopsquark
- Oct 21st 2007, 03:52 PMDunit0001
Ok I think I figured it all out now, thanks!

- Oct 21st 2007, 04:00 PMtopsquark
You probably should be able to factor this by inspection, or if you can't you can use the ac method. (It's a tutorial, just follow the instructions.)

I'm going to use a slightly disparaged method of factoring: factoring by solution. I will note that this method is typically only used when the quadratic will not factor over the integers. (Meaning that does not factor into two linear factors and , where a, b, c, d, e, f, and g are all integers.)

I am showing you this method because I tend to have a mental block when the coefficient of the term is not 1. When you know the quadratic will factor nicely, you should be able to factor the expression by inspection. As earboth said: Practice! (Tmi)

Consider the equation

You don't know how to factor this, and I'm assuming you aren't going to want to go through completing the square (though I recommend doing just that! ;) ), so use the quadratic formula:

etc.

I get:

and .

Now we need to make factors of this polynomial. Let's start with the because it's the harder one.

First, get rid of the fraction:

Now move the constant to the other side of the equation:

What this means is that is a factor of .

You can find the other factor by doing the same thing with the solution.

-Dan