how to find the root in $x^3-x^2+6x+24$? I'm trying my hand on these types of expressions.  how to find the root in $x^3-x^2+6x+24$ ?
please write any idea you have and easy ways please ! thanks.
 A: Let $f(x) = x^3-x^2+6x+24.$
$f(-2) = \cdots = 0$
Therefore $x + 2$ is a factor of $f(x)$.
Find the quotient by long division.
Factorize the quotient.
A: Per the Rational Root Theorem, if there is a rational root for this polynomial, it must be a divisor of 24, i.e., in $\pm \{ 1, 2, 3, 4, 6, 8, 12 \}$.  Trial and error reveals that $x = -2$ is the only root from this set.  Equivalently, $x + 2$ is a factor of the polynomial.
Using polynomial long division, you can factor the polynomial into $(x + 2)(x^2  - 3x + 12)$.  From there, it's a simple matter of applying the quadratic formula.
If you hadn't been fortunate enough to be given a cubic that happened to have a rational root, you could use a generic root-finding method like the bisection method or Newton's method to find that first real root.
There is a Cubic Formula, but it's rather complex.
A: Using the Rational root theorem you can find one rational root $a$.
Then use Polynomial long division to write $$x^3-x^2+6x+24$$  as $$(x-a)g(x)$$ where $g$ is a  quadratic polynomial.
Note that the other two roots of the above polynomials are the roots of $g$
A: A possible way is to try out some of the small values like -1,-2,1 and 2. That's what I do. Usually it does the trick but I don't expect it to work everytime. In your case -2 is a solution. Guessing one root is very helpful.
$$x^3-x^2+6x+24=0$$
$$\Rightarrow x^3+2x^2-3x^2-6x+12x+24=0$$
$$\Rightarrow x^2(x+2)-3x(x+2)+12(x+2)=0$$
$$\Rightarrow (x+2)(x^2-3x+12)=0$$
The quadratic does not have a real root.
