Question

Help for a medal and fan please?

Answer 1

Factor completely 3x3 + 6x2 - 24x.

Answer 2

@johnweldon1993 can you help?

Answer 3

Hey :)

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

Answer 4

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

Answer 5

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

Answer 6

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?

Answer 7

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

Answer 8

Oh, haha thats what you meant XD

Answer 9

Lol yeah but ehh what can ya do :P

Answer 10

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?

Answer 11

Yes I am!

Answer 12

Sorry typo...

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

Answer 13

Its alright

Answer 14

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 :)

Answer 15

Okay, right.

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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?

Answer 20

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?

Answer 21

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

Answer 22

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?

Answer 23

Well first you have to break them down, right?

Answer 24

How so?

Answer 25

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.

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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?

Answer 31

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!

Answer 32

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!

Answer 33

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

Answer 34

Well, not test, but assignment

Answer 35

Lol either way :P