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

1. anonymous

Factor completely 3x3 + 6x2 - 24x.

2. anonymous

@johnweldon1993 can you help?

3. johnweldon1993

4. anonymous

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

5. anonymous

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

6. 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?

7. johnweldon1993

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

8. anonymous

Oh, haha thats what you meant XD

9. johnweldon1993

Lol yeah but ehh what can ya do :P

10. 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?

11. anonymous

Yes I am!

12. johnweldon1993

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

13. anonymous

Its alright

14. 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 :)

15. anonymous

Okay, right.

16. johnweldon1993

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

17. anonymous

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

18. 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

19. 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?

20. 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?

21. anonymous

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

22. 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?

23. anonymous

Well first you have to break them down, right?

24. johnweldon1993

How so?

25. 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.

26. anonymous

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

27. 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

28. johnweldon1993

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

29. anonymous

30. 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?

31. 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!

32. 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!

33. anonymous

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

34. anonymous

Well, not test, but assignment

35. johnweldon1993

Lol either way :P