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, 02:21 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, 02:33 PMgalactus
Practice is what factoring requires.

Quote:

$\displaystyle 3) x^{4} - 8x^{2} + 16$

How about -4 and -4

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

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

Notice the difference of two squares?

$\displaystyle (x-2)(x+2)(x-2)(x+2)=(x+2)^{2}(x-2)^{2}$ - Oct 21st 2007, 02:47 PMtopsquark
Here's another example: factoring by grouping. Rearrange the terms a little.

$\displaystyle x^2 - x - a - a^2 = (x^2 - a^2) + (-x - a)$

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

$\displaystyle = (x - a)(x + a) -1 \cdot (x + a)$

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

$\displaystyle = ((x - a) - 1)(x + a) = (x - a - 1)(x + a)$

You can use the same trick on 4).

-Dan - Oct 21st 2007, 02:52 PMDunit0001
Ok I think I figured it all out now, thanks!

- Oct 21st 2007, 03:00 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 $\displaystyle ax^2 + bx + c$ does not factor into two linear factors $\displaystyle dx + e$ and $\displaystyle fx + g$, 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 $\displaystyle x^2$ 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

$\displaystyle 5s^2 - 16s + 12 = 0$

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:

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

etc.

I get:

$\displaystyle s = 2$ and $\displaystyle s = \frac{6}{5}$.

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

First, get rid of the fraction:

$\displaystyle 5s = 6$

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

$\displaystyle 5s - 6 = 0$

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

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

-Dan