Incredibly small equation...but cant find answer

Haxx0rm4ster

cuberoot(x) = x-6

I know the answer is 8, but I dont know how to solve it... how embarrassing.
Taking Ap chem and not knowing how to find this out. im just in Algebra 2.

topsquark

Cube both sides:
[tex]x=(x-6)^3=x^3-18x^2+108x-216[/tex]

So
[tex]0=x^3-18x^2+107x-216[/tex].

I think the best, low level, way to try to solve this is to use the "Rational Roots Test." It's a simple enough method once you've learned it. Basically, take any polynomial equation of the form [tex]0=ax^n+bx^{n-1}+...+cx+d[/tex]. Look at the leading coefficient (a), in this case it's 1. Factor 1 into all possibilities: [tex]1= \pm 1[/tex]. Next look at the constant term (d) and factor that: [tex]-216=\pm (1, 2, 3, 4, 6, 8, 9, 12, ...)[/tex]. If any rational roots to the equation exist, they must be of the form: x = (factor of constant term)/(factor of leading term). So make your list and try all the possibilities. If one (or more) works, then you've found a solution. If none of them do then the roots are not rational numbers and you need a more advanced method.

-Dan

Haxx0rm4ster

[tex]0=x^3-18x^2+107x-216[/tex]
tahts pretty much where i got to...except i moved 216 to the other side, took out one X from the right side, but noticed the equation was nonfactorable.


well... from the way you said it, it could have just been easier to plug in numbers into X to find a solution... which is what i did to find that X=8.
Obviously, there must be an equation to do this.


If you know how to do this, could you solve it for me? I mean its not a homework it was just a classwork that we turned in to a substitute teacher... but no one knew how to do it.

I just want to learn step by step.

lightgrav

Once you get the first root, just use "long division" :
divide the " x^3 - 18x^2 +107x -216 " , treating the x-powers like place value,
by the " x-8 " root . I get " x^2 - 10x + 27 " .
So, now what are the other roots?
They're not integers, since -3 + -9 is not -10 ... (-3*-9 = 27) but not far off.

topsquark

Actually, there isn't an equation. Well, for cubic and quartic equations there sort of is, but they're rather nastier than the quadratic equation. And if the degree of the polynomial is 5 or larger, you're pretty much stuck with numerical approximation in general.

If you're feeling up to some algebra, look up "Cardano's Method" for solving cubic equations. There's a ton of algebra involved, and the solutions usually need quite a bit of tidying before they simplify for even simple problems, but the process isn't that hard to understand.

-Dan