Solve by Factoring A. B. and C. And factor (if necessary). Then find the GCF?

4a) x^2 - 2x - 15 = 0

Look at the last term -15. What numbers do you know that when multiplied will give you 15?

What numbers do you know that when multiplied will give you -15?

For my first question, you should have 1 and 15 or 3 and 5.

For my second question, you should then have -1 and 15 or 1 and -15 or -3 and 5 or 3 and -5.

Don't continue until you understand what I just asked. We are looking for the factors of the last term in the equation.

Now, look at those factors of -15. Which set of them adds up to -2?

If you indicated 3 and -5, you are correct. +3 -5 = -2

Those numbers are the key to this problem.

Now, we can factor by using them in binomials (expressions with 2 terms)

(x +3)(x-5) = 0 If you multiply those binomials (FOIL) them, you will find that you get your original equation. So if what you are supposed to do is factor, you are done.

If you are supposed to solve for x, you are almost done.

When you have a bunch of factors multiplied together with the result equal to zero, one of the factors has to be equal to zero. We express that by setting each of our factors to zero.

x+3 = 0 or x-5 = 0 Solve for x in each case

x = -3 or x = 5

4b) 2x^2 - 13x + 20 = 0

Here it is not enough just to list the factors of 20 which are 20 and 1 or 10 and 2 or 5 and 4 or -20 and -1 or -10 and -2 or -5 and -4. When we start looking for the combination that will give us the -13 in the middle term, we have to realize that one of the terms will be multiplied by 2.

Our factoring is going to start looking like this (2x - ?) (x - ??)

Notice that I used the negative signs. Since the 20 is positive, the signs of the factors that produce it have to be the same but since the middle term is negative, those have to also be negative.

Think about that for a few minutes if it is not clear. Try a few simple examples and see what happens.

So, when we start looking for combinations that will give us -13, we can probably eliminate the -1 and -20 because it will be too large. We can also eliminate the -2 and -10 because they are both even numbers so that will not work either. Will -4 and -5 work?

We could try (2x -4)(x-5). Multiplying for the middle term we have (2)(-5) + (1)(-4) = -14.

Are we ready to say the set doesn't work? Not until we try them in the other order.

(2x-5)(x-4) Then we have (2)(-4) + (1)(-5) = -8 -5 = -13

So our factoring gives us

(2x-5)(x-4) = 0

The rest is just like the first example.

Give it a try.

5) When you factor x^2 - 4, does it have (x+2) as a factor? (x+2) cannot be factored any more.

When you factor x^2 + 6x +9, does it have any of the same factors as x^2 + x -6?

That is what this problem is asking.