RAM231
  • RAM231
I can't seem to factor this equation out....
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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RAM231
  • RAM231
\[2x ^{2}-2x-12\]
welshfella
  • 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
  • RAM231
whhhhhhhhaaaaaaaaaaaat?

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welshfella
  • welshfella
lol! don't mock!
RAM231
  • RAM231
im so confused
welshfella
  • welshfella
if 2 numbers are a and b ab = -6 and a + b = -1
welshfella
  • welshfella
how about -3 and 2?
RAM231
  • RAM231
these are my choices though....
welshfella
  • welshfella
-3*2 = -6 and -3 + 2 = -1
RAM231
  • RAM231
that isn't one of my choices though!!!
welshfella
  • 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
  • RAM231
@zepdrix you mind helping too?
zepdrix
  • zepdrix
Ugh they didn't fully factor the option choices :( I guess they wanted you to leave the 2 inside.
welshfella
  • welshfella
do you understand how i got the -3 and + 2 ?
RAM231
  • RAM231
yes.
welshfella
  • welshfella
good
RAM231
  • RAM231
@zepdrix what do you mean by they wanted me to leave the 2 inside?
zepdrix
  • 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
  • zepdrix
Confused? :U
welshfella
  • welshfella
these are more difficult to factor if the coeffiecient of x^2 is > 1.
zepdrix
  • zepdrix
Very true :( You have to resort to grouping.
RAM231
  • 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
  • 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
  • RAM231
yeah. its basically like the pathagoream therom or whatever... right?
zepdrix
  • zepdrix
No no no :)
RAM231
  • RAM231
oh. its kinda in that format with the variables
zepdrix
  • 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
  • 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
  • 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
  • zepdrix
ok with that step? :U
RAM231
  • RAM231
yeah. Im with you.
zepdrix
  • 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
  • RAM231
2x
zepdrix
  • zepdrix
Hmm, no. The last term doesn't have an x in it.
zepdrix
  • zepdrix
A 2 is fine, I think we can take a larger amount than a 2 out though.
RAM231
  • RAM231
maybe 6?
zepdrix
  • zepdrix
Hmm we can't take a 6 out of a 4.
RAM231
  • RAM231
then 3(: or 4?
zepdrix
  • 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
  • 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
  • RAM231
wouldn't taking 4 out of 12 be 8 though?
zepdrix
  • zepdrix
When I say "taking out", what I mean is "dividing out"
RAM231
  • RAM231
oh okay. gotcha
zepdrix
  • zepdrix
Do you see anything that they have in common?\[\large\rm [2x\color{orangered}{(x-3)}+4\color{orangered}{(x-3)}]\]
RAM231
  • RAM231
the variable x and the -3... so the ones you highlighted in red
zepdrix
  • 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
  • RAM231
2x+4
zepdrix
  • 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
  • zepdrix
Which of course is the same thing as\[\large\rm (2x+4)(x-3)\]
RAM231
  • RAM231
oh cool! Thanks! I got it right! Thanks @zepdrix are you in college or something? lol
zepdrix
  • zepdrix
Yes, hopefully I'll graduate soon :) Gettin' too old for this school'n stuff lol
welshfella
  • welshfella
@RAM231 - if you would like to see another example its on the following short titorial http://openstudy.com/users/welshfella#/updates/55c0a151e4b01850ec7ff57d
welshfella
  • welshfella
* tutorial
RAM231
  • RAM231
xD bahaha welcome to my world. Ive still got this upcoming year, and 4 years of college/grad school
welshfella
  • welshfella
Mine was over many years ago...
welshfella
  • welshfella
good luck

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