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anonymous
 one year ago
Help for a medal and fan please?
anonymous
 one year ago
Help for a medal and fan please?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Factor completely 3x3 + 6x2  24x.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@johnweldon1993 can you help?

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1Hey :) Wel first off...notice anything about this expression? Common things I mean?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well I noticed there is a cubed and a squared this time

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Like most of my questions didn't have a cubed exponent

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1Well yes that's true lol, but also dont let that deter you...this will get easy very quick! \[\large 3x^3 + 6x^2  24x\] Well first thing I notice...is that each term has at least 1 'x' right?

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1And another thing I notice is that all these numbers are divisible by 3 right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, haha thats what you meant XD

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1Lol yeah but ehh what can ya do :P

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1So! if each of these numbers is divisible by 3...we can factor it out \[\large 3x^2 + 6x^2  24x\] will then become \[\large 3(x^2 + 2x^2  8x)\] still with me?

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1Sorry typo... \[\large 3(x^3 + 2x^2  8x)\]

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1But okay now...since each term ALSO has at least 1 'x' ...we can ALSO factor out an 'x' \[\large 3(x^3 + 2x^2  8x)\] will then become \[\large 3x(x^2 + 2x  8)\] And NOW we have something that can be factored very easily right :)

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1So after we factor what is inside those parenthesis...we get...?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Stop! When you factor, do you multiply or divide? I always get confused by this

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1Well I use factor loosely here I admit Before what we were doing...we were factoring out a common term (With can be thought of as dividing each term we have by that term) Here when I say factor \(\large x^2 + 2x  8\) I mean tell me what that can be broken down into

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oooh, okay. Yeah sorry I was often told different definitions of the word factor so I got confused. So, the 8 can be broken down into 2? Is that what you are looking for?

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1Not quite...well as you can see we have a quadratic, so lets write it as such... \[\large x^2 + 2x  8 = 0\] Now, we can go through the whole quadratic formula and get our roots...OR we can factor since this one is each So... \[\large x^2 + 2x  8 = (x + ?) (x  ?)\] Does that look familiar?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Haha, sorry but I'm still confused by what factoring means XD Sorry for being a pain D:

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1No that's okay...here lets take a tangent for a second If I give you a quadratic expression \[\large ax^2 + bx + c = 0\] How would you solve that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well first you have to break them down, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0By dividing a certain number by a certain number. Sorry I'm not good at explaining so I'll give an example. Earlier, you broke down 3x^3+6x^2−24x into 3(x3+2x2−8x) by dividing them all by 3 since 3 went into all of them.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ugh forgot to put the ^ symbol. 3(x^3+2x^28x)

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1Okay I see what you were going for Not quite That process was a way of simplifying our "hard" expression into an easier one...what we are doing now is the next step now that we have an easy expression

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1SO, what I gonna do is a quick little refresher about factoring :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wow, I am more clueless about this then I thought I was :O

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1If we have a quadratic expression \[\large ax^2 + bx+ c = 0\] We can solve this in 3 different ways...but I will focus on the basic one first The quadratic equation \[\large \frac{b \pm \sqrt{b^2  4ac}}{2a}\] does that look familiar?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oooh, yeah I remember this. I can actually finish it from here, I remember what to do from here. Thank you for your help and time!

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1There ya go :) yeah forget the word "factor" for now...it'll come to you a little later on :) Lemme know what you get for an answer!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, I'll also let you know what I get on my test!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, not test, but assignment

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1Lol either way :P
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