f(x) = -2x^3 - 5x^2 - 6x + 4 ?

What are the zeros of this equation

Question

f(x) = -2x^3 - 5x^2 - 6x + 4 ? 

What are the zeros of this equation

anonymous · Answer

first by trail nd error find a zero...

lgbasallote · Answer

ooohhh sir @eliassaab in action...this is very valuable *_*

*popcorn*

anonymous · Answer

There are two complex roots and one real root. Try graphing.

lgbasallote · Answer

is there a way to know that? like a formula??

anonymous · Answer

no formula first find one zero by trail nd error

anonymous · Answer

There is a general formula for cubics equations, see http://en.wikipedia.org/wiki/Cubic_function

anonymous · Answer

I used the Rational Root Theorem and Descartes' Rule of Signs. But I can't find that one positive zero.

anonymous · Answer

datz y i said try de ec way....first trail nd error den take de factor....den divide u'll get a quadratic polynomial, den split de middle term to find de odr 2 zeroes

anonymous · Answer

Wait I got it thanks eliassab

anonymous · Answer

Here it is
$$x=\frac{1}{6}
   \left(-5-\frac{11}{\sqrt[3]{361+6
   \sqrt{3657}}}+\sqrt[3]{361+6
   \sqrt{3657}}\right)

$$

anonymous · Answer

$$
x\approx 0.458932
$$