Re: solve polynomial for x?

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Posted by T.Gracken on September 18, 2002 at 13:12:54:

In Reply to: Re: solve polynomial for x? posted by colin on September 18, 2002 at 10:45:57:

: : (6+x)^3/(4+x)^2 = 27
: : (6+x)^3 = 27(4+x)^2
: : x^3 - 9x^2 -108x -216 = 0
: : x(x^2 - 9x -108) = 216
: : ...that's as far as I can go;
: : the only integer solution is x = -3.

: how did you get x = -3 from that?

You could approach this as follows...

using Mr.B's result of x^3 - 9x^2 -108x -216 = 0 (from above)

the rational root theorem can lead you to the solution x=-3 (I don't know how mr. B came up with it if he used something else)

then, since x = -3 is a solution, (x + 3) must be a factor of x^3 - 9x^2 -108x -216 = 0 (zero factor theorem)

so use your favorite method of division to rewrite the expression (i used synthetic division) as

(x + 3)(x2 - 12x - 72) = 0

and you can use the quadratic equation to come up with the other two roots.

so, x = -3, or x = 6 + 3*sqrt(6), or x = 6 - 3*sqrt(6).

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