Expert: Josh - 10/13/2009

Question
i need help on this problem.dont understand where to start
x^3+6x^2-4x-24=o
finding the real number solution of the equation

Answer
Alexis,

This question basically asks us to find all possible values of x which make the equation true. There are at most 3 unique solutions (also called "zeros" or "roots") because this is a third order polynomial.

Factorization is the key to this problem. To get anywhere, we need to make these observations.

Observe that we may rearrange the equation as
x^3-4x + 6x^2-24 = 0
Then, we can factorize the 1st and 2nd term, and separately for the 3rd and 4th term.
x(x^2-4) + 6(x^2-4) = 0
Now, since the quadratic factor (x^2-4) is repeated, we can pull this out the front and use it as a common factor. Therefore,
(x^2-4)(x+6)=0  ...[*]
Now, recognize that x^2-4 is a "difference of square" expression. In general, x^2-a^2 = (x+a)(x-a). So, we can decompose [*] further as
(x+2)(x-2)(x+6)=0.

Once we reach this point, finding the solutions is easy. The expression equals zero when either (x+2)=0, (x-2)=0 or (x+6)=0. These occur precisely when x=-2, x=+2 and x=-6. These are the solutions.

Good luck and practice heaps.