By L.D.

Factoring Pt. 1Formula: x2 + bx + c

Problem 11. x2 + 8x + 12

Problem 1

1. x2 + 8x + 12

Now, to solve this problem, we are using reverse F.O.I.L. like we did in the last post, but this time it will look more like (x + x)(x +x) then x(x + x).

Problem 1

1. x2 + 8x + 12

If this variable remains without a coefficient and as a standalone variable then part of our problem is already decided for us. The variable that is like that will always go on the far right of the parenthesis like (x + __ ) (x + __ )Now we need to fill the blank space in the problem at the end of the red. The numbers that fill those spots must always multiply to get the constant and add to get the coefficient. The numbers 2 and 6 always add to get 8 and multiply for twelve, so they are perfect. The full answer is on the next slide. It doesn’t matter what side they choose.

Problem 1 (Answer)

1. (x + 2 ) (x + 6 )

Example Problem to Reinforce Problem 11. x2 + 8x + 12

Example Problem to Reinforce Problem 11. x2 + 8x + 12

Now we know where x will go, and the numbers 6 ad 2 multiply to 12 and add to 8 so our answer is (x + 6)(x + 2). The way to check your answer is to use F.O.I.L. on the problem.

Mini Lesson

It is easier to find the coefficient and constant if you use the cake method on the constant to find its factors. For example, the factors of 12 are 1, 2, 2, & 3. By playing around with these, we can solve our answer easier.

Problem 21. n2 – 5n + 6

Problem 21. n2 – 5n + 6

For this problem, we will treat it a bit differently. To get a negative we need to add two negatives to get -5, we need two negatives so they cancel each other out and can make positive 6. -3 and -2 are perfect so we can but them in a problem on the next slide.

Problem 2 (Answer)1. (n – 3)(n – 2)

Example Problem to Reinforce Problem 2 (with a twist!)

1. x2 – 4x + 3 = 0

Example Problem to Reinforce Problem 2 (with a twist!)

1. x2 – 4x + 3 = 0

Solve it like you would without the zero and then solve for x.

Example Problem to Reinforce Problem 2 (with a twist!)

x2 – 4x + 3 = 0

(x – 3)(x – 1)

x = 3 x = 1

That’s the full problem, so now we can move to the next one!

Problem 31. x(x-3)-24=4

Problem 31. x(x-3)-24=4

First, we need to bring it into the proper formula format, (x2 + bx + c). I will do this on the next slide.

Problem 3x(x-3)-24=4x2 – 3x – 24 = 4 -4 -4x2 – 3x – 28 = 0Now, we will bring the red to the next slide to do the next step.

Problem 3x2 – 3x – 28 = 0

We know where the xs go and we have to do something a bit special to get the other spaces. Our answers are -7 and 4 since -7 + 4 are -3 and they make -28. The last part and answer are on the next page.

Problem 3(x – 7)(x + 4) = 0x = 7 x = 4

Problem 4Find the width (w) of the small shaded space, the area of the full shape is 117 in2, now we will go to the next slide for instructions.

w in

w in + 2 in2 in

Problem 4Now we will find the “area” of the whole thing by doing length times width. The area of the small is 2w and the area of the large is w2 + 2w. Added together, those make 4w + w2 = 117. We will do the problem to match our formula on the next page.

w in

w in + 2 in2 in

Problem 44w + w2 = 117

-117 -117

4w + w2 -117 = 0

(w + 13)(w – 8)

w = -13 w = 8

Since w = 8 is the only plausible answer (-13 in cannot be our width), it is our width and the area of the shaded space is 16.

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