How do I solve these three polynomial equations using factored form?

Question

I was assigned homework tonight of roughly 20 problems... out of these 20 there were 3 total that were worth extra credit. I have a pretty bad teacher, and her words were, "You won't know how to do these, but if you do, you'll get a few points." She never even taught us how to do these 3 and she expects me to get them?

1. x^3 - 12x + 16 = 0

2. x^3 - 12x - 16 = 0

3. x^3 - 3x - 2 = 0

Any help would be greatly appreciated, I'd love to prove her wrong!

Answer 1

First key to factoring: Find all possible factors of your numeric portion. 16 may be split into factors 3 ways:
16 * 1 * 1
8 * 2 * 1
4 * 2 * 2

Second key: Use the sign of the numeric portion to determine how many positive and negative factors you have. A positive numeric portion means that 0 or even number of factors are negative. A negative numeric portion means you have an odd number of negative factors.

Step 1: Always try 2, 1, -1, and -2 first to see where you get. These are easy to plug in and calculate, and are common factors for test questions.

1.) f(x) = x^3 - 12x + 16
f(2) = 8 - 24 + 16 = 0 (factor)
f(1) = 1 - 12 + 16 = 5
f(0) = 16
f(-1) = -1 + 12 + 16 = 27
f(-2) = -8 + 24 + 16 = 32

Since 2 is a factor, let's create a binomial factor that makes sense to along with it. We know that our binomial must have x^2 alone to go with our factor of (x - 2) and get x^3. To get 16, the numeric portion of our new binomial must be 8.
(x^2 + ax - 8)(x - 2) = x^3 - 12x + 16

Now, taking x^3 - 12x + 16 = 0, ax only affects the x^2 term and x term of the trinomial.
ax * x - 2x^2 = 0 (x^2 term)
-8x - 2ax = -12x (x term)

Take these terms, set them up as equations equal to 0, then use both equations against each other to solve for a:
-8x - 2ax + 12x = 0 = 4x - 2ax
4x - 2ax = 0 = ax^2 - 2x^2
4 - 2a = ax - 2x (divide by x)
2x + 4 = ax + 2a
2(x + 2) = a (x + 2)
a = 2

This gives us a factored equation of:
(x^2 + 2x - 8)(x - 2)

Now, let's try to factor x^2 + 2x - 8.
8's factors are 1 and 8 (can't add up to 2) and 2 and 4 (4 + -2 = 2, that works).
Thus, x^2 + 2x - 8 = (x + 4)(x - 2)

Our final factored equation is:
x^3 - 12x + 16 = (x + 4)(x - 2)(x - 2)

Check:
(x + 4)(x - 2)(x - 2) = x^3 + 4x^2 - 2x^2 - 2x^2 - 8x - 8x + 4x + 16 = x^3 - 12x + 16 (check!)

That should help you do the other two.