# 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.

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x=-2 is a root of the cubic. – GTX OC Apr 5 '14 at 16:24
You could use en.wikipedia.org/wiki/Polynomial_remainder_theorem – kingW3 Apr 5 '14 at 16:25

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.

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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.

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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 $g$

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can u please use it so i can see – jackcall Apr 5 '14 at 16:28

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.

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