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anonymous
 5 years ago
Factor Completely:
8x^3+12x^210x15
anonymous
 5 years ago
Factor Completely: 8x^3+12x^210x15

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Everything after the \[8x^3\] I could do now, but would you factor the above like so? (2x)^3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I tried going the right side by itself, but no divisors of 180 = 10. The closest was 10*8

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think you're going to have to factor into three linear equations. ie, (2xa)(2xb)(2xc)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yech, I will try that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That left me with \[2x(4x^2+6x5)15\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Trying to solve that too

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But the 4x^2+65 won't factor either.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm sorry, 4x^2+6x5 I multiplied 5*4 and got 20. I put in the negative sign and got 20. The only things that go into that are 10*2 and 5*4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Neither of those, no matter how you switch them around, equals +6

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Unless the simplest factored form of the top problem is: \[(2x15)(4x^2+6x5)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And that can't be right either. I just started long multiplication, and I already got 75 and that isn't 15

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, that's definitely not right.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Maybe you just can't factor it and it is a trick question?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think you can, I just don't remember how to do 3rd power equations. Still plugging and chugging. ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0He does give us equations that have no solutions all of the time, among ones that look like they don't have solutions but they really do. Ugh, I am going to screw this final up so bad :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I found something on the net saying this is a cubic function and not a quadratic function

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That's annoying. Have you had other problems that factor out to three linear equations? ic, (ax+d)(bx+e)(cx+f)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right, that is true. Quadratic should have a highest power of x^2.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Never heard of it before. But the problem I looked up had something similar and the answered turned into: \[\pm i \sqrt{?}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a 3 was supposed to be in the square root box

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nvm, that was solving for x.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[i=\sqrt{1}\] And I doubt that is what you are working with!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No matter how I break up the problem, nothing factors anywhere other than: \[2x(4x^2+6x5)15\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The next 2 problems after this are easy, so it makes me feel like I should know how to do this.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0look 3/4 of the page down here: http://salinesports.org/mr_frederick/Algebra%202A/Unit%206/Notes_0605.pdf

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm not very familiar with that method, but it does work.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I found a formula, one sec

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have an answer for you.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The b/3a part is a fraction

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what is y? Look at that link if you can. I'll type out the equation here if you want/rather.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I will let you go on this, I have bothered you for way too long :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It is calling it a depressed cubic

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That is what it says to do

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Our objective is to find a real root of the cubic equation ax^3+bx^2+cx+d=0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But anyway, I am going to skip this, because I need to be working on logs and matricies :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's fine. I'll type out what I did. First group: \[(8x ^{3}+12x ^{2})+(10x  15)\] Then factor: \[4x ^{2}(2x+3)  5(2x+3)\] since the factored part (2x+3) is the same, you pair the multiplying factors and get this: \[(4x ^{2}5)(2x+3)\] Which works, and is in it's simplest form (since 4x^25 does not factor further as an integer).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Welcome  thanks for sticking it out. It was going to drive me crazy! ;)
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