Help for a medal and fan please?

- anonymous

Help for a medal and fan please?

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- anonymous

Factor completely 3x3 + 6x2 - 24x.

- anonymous

@johnweldon1993 can you help?

- johnweldon1993

Hey :)
Wel first off...notice anything about this expression? Common things I mean?

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- anonymous

Well I noticed there is a cubed and a squared this time

- anonymous

Like most of my questions didn't have a cubed exponent

- 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

And another thing I notice is that all these numbers are divisible by 3 right?

- anonymous

Oh, haha thats what you meant XD

- johnweldon1993

Lol yeah but ehh what can ya do :P

- 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

Yes I am!

- johnweldon1993

Sorry typo...
\[\large 3(x^3 + 2x^2 - 8x)\]

- anonymous

Its alright

- 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

Okay, right.

- johnweldon1993

So after we factor what is inside those parenthesis...we get...?

- anonymous

Stop! When you factor, do you multiply or divide? I always get confused by this

- 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

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

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

Haha, sorry but I'm still confused by what factoring means XD Sorry for being a pain D:

- 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

Well first you have to break them down, right?

- johnweldon1993

How so?

- 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

Ugh forgot to put the ^ symbol. 3(x^3+2x^2-8x)

- 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

SO, what I gonna do is a quick little refresher about factoring :)

- anonymous

Wow, I am more clueless about this then I thought I was :O

- 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

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

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

Alright, I'll also let you know what I get on my test!

- anonymous

Well, not test, but assignment

- johnweldon1993

Lol either way :P

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