factoring pt. 2/2 (x^2 + bx + c)

Factoring Pt. 2/2 (x2 + bx + c)

By L.D.

Problem 1 3x2 + 8x + 4

Problem 1 3x2 + 8x + 4

The first thing we have to do is to find what multiplies to get 3x2. I choose 3x and x, so we put them in parenthesis;

(3x + ?)(x + ?)

Problem 1 3x2 + 8x + 4

(3x + ?)(x + ?)

The next step is a bit different from what you are expecting, we will take multiplication problems that make 4 and place them in the question mark’s spots, forgetting about the 8 for now. You will do will differentiate between the problems like 1(4) and 4(1) giving them separate problems. These will be made and solved on the next page.

Pairs that make 4: 2 and 2, 1 and 4, 4 and 1

Problem 11. (3x + 2)(x + 2)

2. (3x + 1)(x + 4)

3. (3x + 4)(x + 1)

Use F.O.I.L. to solve these…

Problem 11. (3x + 2)(x + 2)3x2 + 4 + 8x

2. (3x + 1)(x + 4)3x2 + 4 + 4x

3. (3x + 4)(x + 1)3x2 + 4 + 7x

We will now remember the 8 and our original problem (3x2 + 8x + 4), since the top problem contains an 8 and is exactly the same to our problem, our answer is (3x + 2)(x + 2).

Problem 2 2x2 + 13h - 7

Problem 2 2x2 + 13h – 7

1. For our first step, I will say that to make 2x you need 2x and x.

2. Going to our second step we will say that to get -7 you can use -1 and 7, 1 and -7, 7 and -1, -7 and 1

(2x + -1) (x + 7)

(2x + 1) (x + -7)

(2x + 7) (x +-1)

(2x + -7) (x + 1)

Problem 2 2x2 + 13h – 7

(2x + -1) (x + 7) 2x2 – 7 + 13x

(2x + 1) (x + -7) 2x2 – 7 + -13x

(2x + 7) (x +-1) 2x2 – 7 + 5x

(2x + -7) (x + 1) 2x2 – 7 – 5x

Mini LessonDo you remember how in the last slide, when we had problems like (2x + 7) (x +-1) & (2x + -7) (x + 1) where the negative sign just switched sides? A cheat to answer that is to write answer of the problem for the first one (2x2 – 7 + 5x) and when you are answering the next one you can just add a negative to the coefficient.

Problem 2(2x + -1) (x + 7) is the answer since it multiplies to make our original problem, 2x2 + 13h – 7.

Problem 3 -x2 + x + 20

Problem 3 -x2 + x + 20

The way to get around the negative sign on the first problem is to change it to -1(x2 – x – 20) and solve the problem inside the parenthesis ignoring the -1.

-1(x – 5)(x + 4)

Now you can multiply it all by -1.

(-x + 5)(-x - 4)

This is your final answer.

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