anonymous
  • anonymous
Factor Completely: 8x^3+12x^2-10x-15
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Everything after the \[8x^3\] I could do now, but would you factor the above like so? (2x)^3
anonymous
  • anonymous
\[(2x)^3\] rather
anonymous
  • anonymous
I tried going the right side by itself, but no divisors of 180 = -10. The closest was 10*-8

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anonymous
  • anonymous
I think you're going to have to factor into three linear equations. ie, (2x-a)(2x-b)(2x-c)
anonymous
  • anonymous
Yech, I will try that
anonymous
  • anonymous
That left me with \[2x(4x^2+6x-5)-15\]
anonymous
  • anonymous
Trying to solve that too
anonymous
  • anonymous
But the 4x^2+6-5 won't factor either.
anonymous
  • anonymous
I'm sorry, 4x^2+6x-5 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
  • anonymous
Neither of those, no matter how you switch them around, equals +6
anonymous
  • anonymous
Unless the simplest factored form of the top problem is: \[(2x-15)(4x^2+6x-5)\]
anonymous
  • anonymous
And that can't be right either. I just started long multiplication, and I already got -75 and that isn't 15
anonymous
  • anonymous
Yeah, that's definitely not right.
anonymous
  • anonymous
Maybe you just can't factor it and it is a trick question?
anonymous
  • anonymous
I think you can, I just don't remember how to do 3rd power equations. Still plugging and chugging. ;)
anonymous
  • anonymous
He 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
  • anonymous
I found something on the net saying this is a cubic function and not a quadratic function
anonymous
  • anonymous
That's annoying. Have you had other problems that factor out to three linear equations? ic, (ax+d)(bx+e)(cx+f)?
anonymous
  • anonymous
right, that is true. Quadratic should have a highest power of x^2.
anonymous
  • anonymous
Never heard of it before. But the problem I looked up had something similar and the answered turned into: \[\pm i \sqrt{?}\]
anonymous
  • anonymous
a 3 was supposed to be in the square root box
anonymous
  • anonymous
nvm, that was solving for x.
anonymous
  • anonymous
\[i=\sqrt{-1}\] And I doubt that is what you are working with!
anonymous
  • anonymous
No matter how I break up the problem, nothing factors anywhere other than: \[2x(4x^2+6x-5)-15\]
anonymous
  • anonymous
The next 2 problems after this are easy, so it makes me feel like I should know how to do this.
anonymous
  • anonymous
look 3/4 of the page down here: http://salinesports.org/mr_frederick/Algebra%202A/Unit%206/Notes_0605.pdf
anonymous
  • anonymous
I'm not very familiar with that method, but it does work.
anonymous
  • anonymous
I found a formula, one sec
anonymous
  • anonymous
\[x=y-(b)/(3a)\]
anonymous
  • anonymous
I have an answer for you.
anonymous
  • anonymous
The b/3a part is a fraction
anonymous
  • anonymous
what is y? Look at that link if you can. I'll type out the equation here if you want/rather.
anonymous
  • anonymous
I will let you go on this, I have bothered you for way too long :P
anonymous
  • anonymous
It is calling it a depressed cubic
anonymous
  • anonymous
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anonymous
  • anonymous
That is what it says to do
anonymous
  • anonymous
Our objective is to find a real root of the cubic equation ax^3+bx^2+cx+d=0.
anonymous
  • anonymous
is what it says
anonymous
  • anonymous
But anyway, I am going to skip this, because I need to be working on logs and matricies :P
anonymous
  • anonymous
It'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^2-5 does not factor further as an integer).
anonymous
  • anonymous
Good luck. :)
anonymous
  • anonymous
tyvm!
anonymous
  • anonymous
Welcome - thanks for sticking it out. It was going to drive me crazy! ;)

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