Factoring by Using a Quadratic Pattern; Then solving

[Ash3024]

Hey there! I usually have no problems with Algebra II, but I have come across a problem that I cannot figure out. Right now, we are working on:

Factoring by Using a Quadratic Pattern

The quadratic expression is ax^2+bx+c.

An example of what we have been doing is:

Factor x^4-2x^2-8 *Using temporary subsitution*

1. x^4-2x^2-8 = (x^2)^2-2(x^2)-8 a=x^2
2. x^4-2x^2-8 = a^2-2a-8 substitue a for x^2
3. x^4-2x^2-8 = a^2-2a-8 = (a-4)(a+2) Factored
4. (a-4)(a+2) = (x^2-4)(x^2+2) x^2 is substituted back in
5. (x^2-4)(x^2+2) = (x+2)(x-2)(x^2+2) Factored Completely
6. x = -2, 2, i*(2)^1/2 ((i*(2)^1/2 is i(Squareroot of 2)))

Okay, so the problem I am having trouble with is...

2x^3=5x^2+12x

I know I need to get all the terms on one side, thus I would do this:

2x^3-5x^2-12x=0 ((Subtracting the 5x^2 and 12x is easier than making the leading coefficient negative))

My problem is, how would I make this into a quadratic expression by temporary substitution if there would be 3 terms with x in it, and no constant. Having no constant isn't the problem, I just don't know how to solve it with 3 terms with x in them (Something like x^3+x^2+x). Do you understand?
I have like three more problems like this and I'm confused.

Thanks a lot

 

[tkhunny]

2x^3-5x^2-12x=0

You'll kick yourself.

2x^3-5x^2-12x = x(2x^2-5x-12)

Work on the part in the parentheses.

 

[Ash3024]

Oh gosh! I can be stupid like that sometimes. I was trying to complicate it. Ok, then. I got that problem and another solved, but I have two more which aren't working. Both are with the same concept:

x^3+3x^2-4x-12=0
OK! So you can't factor out x in all of the terms, so I tried doing this:

x(x^2+3x-4)-12=0
I continued working it like before, and got:

x(x+4)(x-1)-12 and so the answers for x would be:
x=0,-4,1, and what about that -12?

The other:

Forget about it, because I figured it out as I was typing it up to explain it to you.

So, does anyone know how to solve the first problem I have on here, using the concept of the problem I had in the original post?

 

[tkhunny]

x(x+4)(x-1)-12 and so the answers for x would be:
x=0,-4,1, and what about that -12?

You are missing a very important point.

If A*B = 0, then either A = 0 or B = 0. This is the important premise. There is ONLY multiplication there.

A*B + C = 0 is no good. We cannot tell anything about A or B in this structure.

You can't lock yourself into one way of thinking. Keep ALL your tools handy ALL the time.

x^3+3x^2-4x-12 = x^2*(x+3) - 4(x+3) = (x^2 - 4)(x+3) = (x+2)(x-2)(x+3)

Now one can find solutions.

 

[Ash3024]

Thanks much...