Fun with Algebra

[XXXconanXXX]

Let me start off bluntly: I've never been good at math. I'm here to consult the good people of TheHelper (Fitting name) for some help with my algebra final review in college. So here goes...

Solving cubic functions. I've been given x³-6x²-20=0
Can I put 20 on the right, pull out an x from the rest, and set both the x and the rest equal to 20 then use the quadratic formula/complete the square for the remaining quadratic? Written out I mean like this:
x(x²-6x) = 20
x = 20 x² - 6x = 20
Then figure the two roots for the remaining quadratic? If not how do I go about solving this?

 

[Sevion]

No, that won't work.

Proof:

Spoiler

Factored formula: (X+2)(X-3)(X+6)=0
Solutions: X=-2,3,-6

Expanded form: X³+5X²-12X-36=0

1. X³+5X²-12X=36, Addition Property of Equality
2. X(X²+5X-12)=36, Factor
3. X=36 (not a solution), X²+5X-12=36 (Setting each side to 36)
4. (-5+-Sqrt(25-4*1*(-48)))/2=-2.5+-14.731, Quadratic Formula (not a solution)

I think the easiest way is to factor this thing using factoring.

An easy way is to find factors of 36.

+-1
+-2
+-3
+-4
+-6
+-9
+-12
+-13
+-36

Now, we just evaluate each one.

F(3)=(3)³+5(3)²-12(3)-36=>27+45-36-36=>72-72=0

Now we have (X-3) as a factor, or X=3.

Factor that out of the original form: (X-3)(X²+8X+12)

Now we CAN use the Quadratic Formula and you'll find that it evaluates to X=-2,-6 making the last factors (X+2) and (X+6).

This is mostly hit-and-miss until you find the first factor. Once you find the first factor, you can use polynomial long division until it's a quadratic. Once it's a quadratic, you can drop into the familiar path of the Quadratic formula (or just factor if that's easier).

 

[Darthfett]

http://www.wikihow.com/Factor-a-Cubic-Polynomial

This gives a list of the Steps Sevion took to find the first factor.

 

[Vellu]

x³-6x²-20=0
For this kind of stuff, I just punch those numbers into my calculator and it will give me the correct answer right away.
I'm studying the "longer" maths as we say it around here ATM and we've only seen 2nd degree polynoms so far. Right now we're doing some limit (lim) stuff - it's difficult sometimes.

 

[Darthfett]

Please look at the last post date before you post next time, amit28it.