Thread: Factoring?

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

Practice is what factoring requires.

What two numbers when multiplied equal 16 and when added equal -8

How about -4 and -4





Notice the difference of two squares?

Originally Posted by Dunit0001

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

Here's another example: factoring by grouping. Rearrange the terms a little.


Note that the first group factors and the second group factors:


Now both the first and the last term have a common factor, x + a, so factor this:


You can use the same trick on 4).

-Dan

Ok I think I figured it all out now, thanks!

Originally Posted by Dunit0001

5) 5s^2 - 16s + 12

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!

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