anonymous
  • anonymous
Help for a medal and fan please?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Factor completely 3x3 + 6x2 - 24x.
anonymous
  • anonymous
@johnweldon1993 can you help?
johnweldon1993
  • johnweldon1993
Hey :) Wel first off...notice anything about this expression? Common things I mean?

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anonymous
  • anonymous
Well I noticed there is a cubed and a squared this time
anonymous
  • anonymous
Like most of my questions didn't have a cubed exponent
johnweldon1993
  • johnweldon1993
Well 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
  • johnweldon1993
And another thing I notice is that all these numbers are divisible by 3 right?
anonymous
  • anonymous
Oh, haha thats what you meant XD
johnweldon1993
  • johnweldon1993
Lol yeah but ehh what can ya do :P
johnweldon1993
  • johnweldon1993
So! 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?
anonymous
  • anonymous
Yes I am!
johnweldon1993
  • johnweldon1993
Sorry typo... \[\large 3(x^3 + 2x^2 - 8x)\]
anonymous
  • anonymous
Its alright
johnweldon1993
  • johnweldon1993
But 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 :)
anonymous
  • anonymous
Okay, right.
johnweldon1993
  • johnweldon1993
So after we factor what is inside those parenthesis...we get...?
anonymous
  • anonymous
Stop! When you factor, do you multiply or divide? I always get confused by this
johnweldon1993
  • johnweldon1993
Well 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
  • anonymous
Oooh, 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
  • johnweldon1993
Not 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
  • anonymous
Haha, sorry but I'm still confused by what factoring means XD Sorry for being a pain D:
johnweldon1993
  • johnweldon1993
No 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
  • anonymous
Well first you have to break them down, right?
johnweldon1993
  • johnweldon1993
How so?
anonymous
  • anonymous
By 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
  • anonymous
Ugh forgot to put the ^ symbol. 3(x^3+2x^2-8x)
johnweldon1993
  • johnweldon1993
Okay 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
  • johnweldon1993
SO, what I gonna do is a quick little refresher about factoring :)
anonymous
  • anonymous
Wow, I am more clueless about this then I thought I was :O
johnweldon1993
  • johnweldon1993
If 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
  • anonymous
Oooh, 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
  • johnweldon1993
There 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
  • anonymous
Alright, I'll also let you know what I get on my test!
anonymous
  • anonymous
Well, not test, but assignment
johnweldon1993
  • johnweldon1993
Lol either way :P

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