#1
Maybe I'm just tired, but I've worked through all my homework without much difficulty, but can't find out how to reach the answer here. I have the answer, but don't know how to get it.

It's college algebra and it's just factoring.

2x^3 + 6x^2 - 8x - 24

I factored out the 2
then tried grouping, which didn't work.
#3
2(x^3+3x^2-4x-12)

I think that's all you can do... that's if I remember my math right, but from what I remember, there isn't any other way to continue factoring since there is no more like values...
#7
Quote by \m/NikkiRyan\m/
2(x^3+3x^2-4x-12)

I think that's all you can do... that's if I remember my math right, but from what I remember, there isn't any other way to continue factoring since there is no more like values...


u can factor that down too, if i remember it right...

i wanna say its 2(x^2-4)(x+3)
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deanwad could be right, actually.


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Last edited by deanwad at Aug 27, 2008,
#9
Quote by APE20
Maybe I'm just tired, but I've worked through all my homework without much difficulty, but can't find out how to reach the answer here. I have the answer, but don't know how to get it.

It's college algebra and it's just factoring.

2x^3 + 6x^2 - 8x - 24

I factored out the 2
then tried grouping, which didn't work.

2+3+6+2+8=22
22-24=-2.
#10
if this is equal to zero, the best you're going to get is 2(x^3+3x^2-4x-12)=0.

It doesn't go any farther.
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#11
Quote by the_man101
add like terms?

2x2x2=8x
6x6=36x

8x+36x=44x
44x-8x=36x-24
idk that would be my best guess.



well yea cuz once you subtract 2 from three you get one, and if any of you remember every variable has an "imaginary" 1 exponent

so after that you combine the like terms.....dont take my word though!!!
#13
Quote by killer puppy
i think you subtract the exponent 2 from 3 and then combine the like terms...



Can't do that. You can divide by x, but that -12 is there after factoring out the two. I'm pretty sure all you can do is divide everything by 2.

EDIT: Nevermind. Forgot about synthetic division. Give that a try.
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#14
Quote by deanwad
u can factor that down too, if i remember it right...

i wanna say its 2(x^2-4)(x+3)


you need to factor x^2-4 soooo

2(x+2)(x-2)(x+3) is final answer TS
#15
2 (x + 2) (x - 2) (x + 3)


/8thgrademath


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Last edited by MTVget0FFtheAIR at Aug 27, 2008,
#16
Common factor a 2x out of the first three terms to get the following;

2x(x^2 + 3x - 4) - 24

Now factor the quadratic in the brackets into two binomials to get the following;

2x(x+4)(x-1) - 24

That's as far as this will go
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#18
I'm wrong. sik209 factored it into three binomials with a coefficient - which is correct
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#19
you guys, check the problem i posted. if it works then it is right (i may have messed up in typing)


2(x+2)(x-2)(x+3)

EDIT: lol, took the pit 20 posts to get the answer to one algebra problem
Last edited by sik209 at Aug 27, 2008,
#20
Quote by MTVget0FFtheAIR
2 (x + 2) (x - 2) (x + 3)


/8thgrademath

Oh yeah of course

There are kids at my engineering school that still don't know how to factor completely.
#21
Sik and MTV were correct. I'm just tired and forgot how to factor a perfect square binomial. Like forgetting how to spell your brothers name, but in math.
#22
Quote by SPrestigious
Common factor a 2x out of the first three terms to get the following;

2x(x^2 + 3x - 4) - 24

Now factor the quadratic in the brackets into two binomials to get the following;

2x(x+4)(x-1) - 24

That's as far as this will go



Nah you can't do that first step as it's all one function.