Question

I can't seem to factor this equation out....

Answer 1

\[2x ^{2}-2x-12\]

Answer 2

first divide by 2

2(x^2 - x - 6)

now you need to numbers whose product is -6 aaaaand whose sum is -1.

Answer 3

whhhhhhhhaaaaaaaaaaaat?

Answer 4

lol!
don't mock!

Answer 5

im so confused

Answer 6

if 2 numbers are a and b
ab = -6 and a + b = -1

Answer 7

how about -3 and 2?

Answer 8

these are my choices though....

Answer 9

-3*2 = -6
and -3 + 2 = -1

Answer 10

that isn't one of my choices though!!!

Answer 11

you can still get using my way
x^2 - x - 6 = (x + 2)(x - 3)

2(x^2 - x - 6) = (2x + 4)(x - 3)

Answer 12

@zepdrix you mind helping too?

Answer 13

Ugh they didn't fully factor the option choices :(
I guess they wanted you to leave the 2 inside.

Answer 14

do you understand how i got the -3 and + 2 ?

Answer 15

yes.

Answer 16

good

Answer 17

@zepdrix what do you mean by they wanted me to leave the 2 inside?

Answer 18

The answer choices are not fully factored, unfortunately :(
So you would probably be better off not taking a 2 out of everything as a first step.\[\large\rm 2x^2-2x-12\]
ac = -24
b=-2

So ummm, factors of -24 hmm
-6*4 = -24
-6 + 4 = -2
ah ha! we found the right ones!
So we want to be clever and rewrite our -2x as the sum of -6x and 4x.\[\large\rm 2x^2\color{orangered}{-2x}-12\]\[\large\rm 2x^2\color{orangered}{-6x+4x}-12\]And then proceed to factor by grouping.

Answer 19

Confused? :U

Answer 20

these are more difficult to factor if the coeffiecient of x^2 is > 1.

Answer 21

Very true :(
You have to resort to grouping.

Answer 22

that makes much more sense to me.... @zepdrix how would you group them? by like variables and stuff? and where did you get the c from?

Answer 23

\[\large\rm ax^2+bx+c\]And we have this equation,\[\large\rm 2x^2-2x-12\]The negative part of each number, not the formula,
so for example b=-2, not 2!
Think of your equation like this if it helps,\[\large\rm 2x^2+-2x+-12\]Then maybe it's easier to see that your
a=2
b=-2
c=-12

Answer 24

yeah. its basically like the pathagoream therom or whatever... right?

Answer 25

No no no :)

Answer 26

oh. its kinda in that format with the variables

Answer 27

Pythagorean relates the three numbers together by squaring each.
Here, we instead have a weird method of multiplying the numbers together.

We multiply a and c, and then look at factors of that number.
We want the factors which add to the middle number.

In this problem we have a ton of options for our factors of ac.
\(\large\rm ac=2\cdot(-12)=-24\)

-8*3 = -24
But -8 + 3 is not equal to -2 (our b value).
So these are not the correct factors.

-2*12 = -24
but -2 + 12 is not equal to -2.

-4*6 = -24
But -4 + 6 is 2, not -2 like we want.

-6*4 = -24
and -6 + 4 = -2, yay.

It's a lot of trial and error with this process.

Answer 28

Once you've found your two magical numbers,
you rewrite the middle in terms of those numbers.
and then we can proceed to finish up the problem in this way:

Answer 29

\[\large\rm \color{royalblue}{2x^2-6x}+4x-12\]To finish with the factoring, hmm let's see.
You want to deal with these `two at a time`.


The first two terms both have x's in them,
so we can pull out an x.
They also have a 2 as a factor, ya?
So let's pull 2x out of each of the first two terms.\[\large\rm \color{royalblue}{2x(x-3)}+4x-12\]

Answer 30

ok with that step? :U

Answer 31

yeah. Im with you.

Answer 32

Let's look at the other "pair",\[\large\rm 2x(x-3)+\color{royalblue}{4x-12}\]What can we take out of each of these blue guys?

Answer 33

2x

Answer 34

Hmm, no.
The last term doesn't have an x in it.

Answer 35

A 2 is fine, I think we can take a larger amount than a 2 out though.

Answer 36

maybe 6?

Answer 37

Hmm we can't take a 6 out of a 4.

Answer 38

then 3(: or 4?

Answer 39

We can take a 4 out of a 4,
and we can take a 4 out of a 12,
ok ya that works :o

Taking the 4 out of 4x leaves us with x as the first term in the brackets,
and taking 4 out of 12 leaves us with 3 as the other value.\[\large\rm 2x(x-3)+\color{royalblue}{4x-12}\]\[\large\rm 2x(x-3)+\color{royalblue}{4(x-3)}\]

Answer 40

\[\large\rm 2x(x-3)+4(x-3)\]This next step is a little tough.
Let's artificially place some square brackets around the outside,\[\large\rm [2x(x-3)+4(x-3)]\]From this point, we want to factor again.
We want to figure out what each term has in common, and pull it all the way outside of the square brackets.

Answer 41

wouldn't taking 4 out of 12 be 8 though?

Answer 42

When I say "taking out", what I mean is "dividing out"

Answer 43

oh okay. gotcha

Answer 44

Do you see anything that they have in common?\[\large\rm [2x\color{orangered}{(x-3)}+4\color{orangered}{(x-3)}]\]

Answer 45

the variable x and the -3... so the ones you highlighted in red

Answer 46

So we'll factor that out of each term, bringing it to the outside of the square brackets.\[\large\rm [2x\color{orangered}{(x-3)}+4\color{orangered}{(x-3)}]\]\[\large\rm \color{orangered}{(x-3)}[\qquad?\qquad+\qquad?\qquad]\]What are we left with in the square brackets when we do this?

Answer 47

2x+4

Answer 48

\[\large\rm \color{orangered}{(x-3)}[2x+4]\]Ok good good good.
And there is our final result!
We can change the square brackets to round brackets so it matches our answer choices a little better,\[\large\rm (x-3)(2x+4)\]

Answer 49

Which of course is the same thing as\[\large\rm (2x+4)(x-3)\]

Answer 50

oh cool! Thanks! I got it right! Thanks @zepdrix are you in college or something? lol

Answer 51

Yes, hopefully I'll graduate soon :)
Gettin' too old for this school'n stuff lol

Answer 52

@RAM231 - if you would like to see another example its on the following short titorial

http://openstudy.com/users/welshfella#/updates/55c0a151e4b01850ec7ff57d

Answer 53

* tutorial

Answer 54

xD bahaha welcome to my world. Ive still got this upcoming year, and 4 years of college/grad school

Answer 55

Mine was over many years ago...

Answer 56

good luck