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

- RAM231

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

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- RAM231

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

- welshfella

first divide by 2
2(x^2 - x - 6)
now you need to numbers whose product is -6 aaaaand whose sum is -1.

- RAM231

whhhhhhhhaaaaaaaaaaaat?

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## More answers

- welshfella

lol!
don't mock!

- RAM231

im so confused

- welshfella

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

- welshfella

how about -3 and 2?

- RAM231

these are my choices though....

##### 1 Attachment

- welshfella

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

- RAM231

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

- welshfella

you can still get using my way
x^2 - x - 6 = (x + 2)(x - 3)
2(x^2 - x - 6) = (2x + 4)(x - 3)

- RAM231

@zepdrix you mind helping too?

- zepdrix

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

- welshfella

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

- RAM231

yes.

- welshfella

good

- RAM231

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

- zepdrix

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.

- zepdrix

Confused? :U

- welshfella

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

- zepdrix

Very true :(
You have to resort to grouping.

- RAM231

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?

- zepdrix

\[\large\rm ax^2+bx+c\]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

- RAM231

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

- zepdrix

No no no :)

- RAM231

oh. its kinda in that format with the variables

- zepdrix

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.

- zepdrix

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:

- zepdrix

\[\large\rm \color{royalblue}{2x^2-6x}+4x-12\]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.\[\large\rm \color{royalblue}{2x(x-3)}+4x-12\]

- zepdrix

ok with that step? :U

- RAM231

yeah. Im with you.

- zepdrix

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?

- RAM231

2x

- zepdrix

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

- zepdrix

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

- RAM231

maybe 6?

- zepdrix

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

- RAM231

then 3(: or 4?

- zepdrix

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.\[\large\rm 2x(x-3)+\color{royalblue}{4x-12}\]\[\large\rm 2x(x-3)+\color{royalblue}{4(x-3)}\]

- zepdrix

\[\large\rm 2x(x-3)+4(x-3)\]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.

- RAM231

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

- zepdrix

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

- RAM231

oh okay. gotcha

- zepdrix

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

- RAM231

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

- zepdrix

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?

- RAM231

2x+4

- zepdrix

\[\large\rm \color{orangered}{(x-3)}[2x+4]\]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,\[\large\rm (x-3)(2x+4)\]

- zepdrix

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

- RAM231

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

- zepdrix

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

- welshfella

@RAM231 - if you would like to see another example its on the following short titorial
http://openstudy.com/users/welshfella#/updates/55c0a151e4b01850ec7ff57d

- welshfella

* tutorial

- RAM231

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

- welshfella

Mine was over many years ago...

- welshfella

good luck

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