# Factoring?

• Oct 21st 2007, 02:21 PM
Dunit0001
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
• Oct 21st 2007, 02:33 PM
galactus
Practice is what factoring requires.

Quote:

$3) x^{4} - 8x^{2} + 16$
What two numbers when multiplied equal 16 and when added equal -8

$x^{4}-4x^{2}-4x^{2}+16$

$x^{2}(x^{2}-4)-4(x^{2}-4)$

Notice the difference of two squares?

$(x-2)(x+2)(x-2)(x+2)=(x+2)^{2}(x-2)^{2}$
• Oct 21st 2007, 02:47 PM
topsquark
Quote:

Originally Posted by Dunit0001
1) x^2 - x - a - a^2

Here's another example: factoring by grouping. Rearrange the terms a little.
$x^2 - x - a - a^2 = (x^2 - a^2) + (-x - a)$

Note that the first group factors and the second group factors:
$= (x - a)(x + a) -1 \cdot (x + a)$

Now both the first and the last term have a common factor, x + a, so factor this:
$= ((x - a) - 1)(x + a) = (x - a - 1)(x + a)$

You can use the same trick on 4).

-Dan
• Oct 21st 2007, 02:52 PM
Dunit0001
Ok I think I figured it all out now, thanks!
• Oct 21st 2007, 03:00 PM
topsquark
Quote:

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 $ax^2 + bx + c$ does not factor into two linear factors $dx + e$ and $fx + g$, 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 $x^2$ 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
$5s^2 - 16s + 12 = 0$

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:
$s = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(5)(12)}}{2 \cdot 5}$

etc.

I get:
$s = 2$ and $s = \frac{6}{5}$.

Now we need to make factors of this polynomial. Let's start with the $s = \frac{6}{5}$ because it's the harder one.

First, get rid of the fraction:
$5s = 6$

Now move the constant to the other side of the equation:
$5s - 6 = 0$

What this means is that $5s - 6$ is a factor of $5s^2 - 16s + 12 = 0$.

You can find the other factor by doing the same thing with the $s = 2$ solution.

-Dan