Could someone please walk me through this factoring problem?

(x - 14) (x + 2)

*Not much work needed...just need to follow some steps.

FOIL Method: (x ) (x )

Using this as a reference: x^2 + bx + c...which I'm sure you can correspond with the problem you gave.

IF c is positive, your factors are going to both have the same sign depending on b’s sign. Since it is not...then

Then, we need two numbers whose product is -28 and sum is -12. That would have to be -14 and 2...which leads to the answer on the top.

Ask yourself this question: what adds to give me -12 and multiplies to give -28?

the fact that the last number, -28, is negative, tells you that of the two numbers you are looking for, one is negative and ove is positive.

what multiplies to give 28?

1, 28

2, 14

3, 7

if one number is negative, do any of those add to -12?

1-28 = -27

2 - 14 = -12

3 - 7 = -4

so it is 2 and -14

Split the middle factor into the two numbers and go from there:

X^2 - 14X+2X - 28 -------> (X^2 - 14X)+(2X - 28) ---------> X(X-14) + 2(x-14) = (x+2)(x-14)

(x+2)(x-14) is your answer

(some people skip right to here, once they find the two numbers, but wait until you are more confident to do so..)

When you have a quadratic function with roots x1 and x2, the factoring is

y = (x-x1)(x-x2)

You function y = x^2 -12x - 28 have the roots x1 = 14 and x2 = -2

Then the factoring is y = (x-14)(x+2) OK!

Ask yourself, which two numbers MULTIPLY to get -28 and ADD to get a -12?

2 and -14, right?

So put these in factored form:

(x+2)(x-14).

There you have it

Factor

Expression

X^2 - 12X - 28

Result

(X-14)(X+2)

Below is a link to an article on WikiHow that may be able to help you solve this problem.

http://www.wikihow.com/Factor-Second-Degree-Polyno...

Good luck and I hope I helped you!

Well, I know (-14)(2) = -28

and -14+2 = -12

Then you'll write it as:

(x-14)(x+2)

(x-14)(x+2)

Hope that helps. :)