Factor Completely:
8x^3+12x^2-10x-15

- anonymous

Factor Completely:
8x^3+12x^2-10x-15

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- schrodinger

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- anonymous

Everything after the \[8x^3\] I could do now, but would you factor the above like so?
(2x)^3

- anonymous

\[(2x)^3\] rather

- anonymous

I tried going the right side by itself, but no divisors of 180 = -10. The closest was 10*-8

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## More answers

- anonymous

I think you're going to have to factor into three linear equations. ie,
(2x-a)(2x-b)(2x-c)

- anonymous

Yech, I will try that

- anonymous

That left me with \[2x(4x^2+6x-5)-15\]

- anonymous

Trying to solve that too

- anonymous

But the 4x^2+6-5 won't factor either.

- 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

Neither of those, no matter how you switch them around, equals +6

- anonymous

Unless the simplest factored form of the top problem is:
\[(2x-15)(4x^2+6x-5)\]

- anonymous

And that can't be right either. I just started long multiplication, and I already got -75 and that isn't 15

- anonymous

Yeah, that's definitely not right.

- anonymous

Maybe you just can't factor it and it is a trick question?

- anonymous

I think you can, I just don't remember how to do 3rd power equations. Still plugging and chugging. ;)

- 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

I found something on the net saying this is a cubic function and not a quadratic function

- anonymous

That's annoying. Have you had other problems that factor out to three linear equations? ic, (ax+d)(bx+e)(cx+f)?

- anonymous

right, that is true. Quadratic should have a highest power of x^2.

- anonymous

Never heard of it before. But the problem I looked up had something similar and the answered turned into:
\[\pm i \sqrt{?}\]

- anonymous

a 3 was supposed to be in the square root box

- anonymous

nvm, that was solving for x.

- anonymous

\[i=\sqrt{-1}\]
And I doubt that is what you are working with!

- anonymous

No matter how I break up the problem, nothing factors anywhere other than:
\[2x(4x^2+6x-5)-15\]

- anonymous

The next 2 problems after this are easy, so it makes me feel like I should know how to do this.

- anonymous

look 3/4 of the page down here:
http://salinesports.org/mr_frederick/Algebra%202A/Unit%206/Notes_0605.pdf

- anonymous

I'm not very familiar with that method, but it does work.

- anonymous

I found a formula, one sec

- anonymous

\[x=y-(b)/(3a)\]

- anonymous

I have an answer for you.

- anonymous

The b/3a part is a fraction

- anonymous

what is y? Look at that link if you can. I'll type out the equation here if you want/rather.

- anonymous

I will let you go on this, I have bothered you for way too long :P

- anonymous

It is calling it a depressed cubic

- anonymous

##### 1 Attachment

- anonymous

That is what it says to do

- anonymous

Our objective is to find a real root of the cubic equation
ax^3+bx^2+cx+d=0.

- anonymous

is what it says

- anonymous

But anyway, I am going to skip this, because I need to be working on logs and matricies :P

- 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

Good luck. :)

- anonymous

tyvm!

- anonymous

Welcome - thanks for sticking it out. It was going to drive me crazy! ;)

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