Factoring Polynomials SUM AND PRODUCT Because polynomials come in many shapes and sizes, there are several patterns you need to recognize and there are different methods for solving them. We are going to start today with the second easiest form of factoring (the easiest being factoring out the GCF). All of the polynomials that we will be factoring today will be quadratic (or quadratic form), mostly trinomials. Some, towards the end, aren’t polynomials, but we will still be able to factor them. Also, most of them will start with x2. There will be some special cases where the x2 has a coefficient other than one, but we are going to save the bulk of that type until day 2 of factoring. Factoring mainly requires an understanding of how polynomial multiplication works and logic. We are basically going to reason our way through this process.

Factoring Polynomials: Type 1: Quadratic Trinomials with a Leading coefficient of 1 Here is the actual procedure we are going to use to factor quadratic trinomials of the form x2 + bx + c. This is going to work for us because we know what happens when we multiply two linear binomials. The x2 term comes from x*x. These x’s come from the first thing we multiply together. That means that our factors have to start (x …) (x …). The numbers that go after the x have to multiply together to give us “c”, which is why we need to find factors of “c”. We are going to include the sign when we think of factors. It will make a difference! We also know that the “bx” term comes from, in essence, adding the factors of “c”. So, if c has multiple factor pairs, we need to find the one pair that will add to b. Remember to incorporate the sign of the factors of c. The final step is to always check to make sure everything works right. It is really easy to pick the wrong factors or the wrong sign on the factors. - 3 - 3 = 2 Binomial Factors

x2 + 8x + 7

x2 – 13x + 40 x2 + 11x + 18

Factoring Quadratic Equations with a = 1 Try these examples: x2 + 7x + 12 2) x2 + 8x + 12 3) x2 + 2x – 3 4) x2 – 6x + 8 5) x2 + x – 12 6) x2 – 3x – 10 x2 – 8x + 15 8) x2 – 3x – 18 9) x2 – 3x + 2 10) x2 – 10x + 21 Examples. I’ll go through the reasoning on some of these, but give the solution to all of them. #1 a) We need to look for factors of +12 that add to 7. This means they both have to be positive. The possible factor pairs are (1, 12), (2, 6), and (3,4). b) Which of these pairs add up to 7? That would be 3 and 4. c) It seems that (x + 3)(x + 4) might be the factors. They add up to 7 and multiply out to 12. d) Time to check (x+3)(x+4) = x2+ 4x + 3x + 12 = x2 + 7x + 12! #2. (x + 6)(x+2) #3 a) We need to look for factors of –3 that add to 2. This means that one is negative and one is positive. Also that the positive one is “bigger.” The possible factor pairs are (1, -3) and (-1, 3). b) Which of these add up to +2? (-1, 3). c) (x – 1) (x+3) looks to be the answer. d) Time to check (x-1)(x+3) = x2 + 3x – x – 3 = x2 + 2x – 3 #4 a) We need to look for factors of +8 that add to –6. This means that they both are negative. The possible factor pairs are (-1, -8), (-2, -4). b) Which of these add up to –6? (-2, -4) c) (x – 2)(x –4) might be the solution. d) Check: (x-2)(x-4) = x2 – 4x – 2x + 8 = x2 – 6x + 8 #5. ( x + 4)(x – 3) #6) (x-5)(x+2) #7) (x – 3)(x-5) #8) ( x – 6)(x + 3) #9) ( x – 2)(x-1) #10) (x – 7)(x-3)