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$\displaystyle 3) x^{4} - 8x^{2} + 16$
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?
$\displaystyle (x-2)(x+2)(x-2)(x+2)=(x+2)^{2}(x-2)^{2}$
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
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!
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