Question

if factorise is not possible for any cubic equation then how to solve

Answer 1

hey friends it also have ohter method to factorise..

example..

f(x)=x3+4x2+x-6=0 [1]

Descartes' Rule of Signs

The signs of the coefficients are:
+ + + -
Therefore there is one positive real root of the equation. We know every cubic has at least one real root, and now we know that for this cubic, the root is positive. 
As every cubic has 3 roots, then, we know immediately that this one has 2 negative real roots or none. 
Just for the record, the signs of the coefficients for f(-x) are:
- + - -
Showing two sign changes and indicating 2 or no negative real roots.

Rational Root theorem

If the equation has rational roots, p/q, then p|-6=1.2.3, and q|1, where "|" means "is a factor of". 
This means any rational roots would be: �(1/1, 2/1, 3/1, or 6/1). 
In this case, we can quickly check that 1 is a root. Because we must use all the factors in our roots, if one factor is 1, then the other two must be �2 or �3, or �1 and �6, if the roots are rational.
We might at this stage realise that the equation can be easily factorised.

Vieta's Root Theorem

α+β+γ=-4
αβγ=6
With one positive root, the possible roots could be -2, -3 and 1, by inspecting the possibilities from the rational root theorem. 
Other possibilities, such as 1, 1, -6 do not fit as Descartes has said there is only one positive real root. Naturally, 1 is easy to check and we find that f(1)=1+4+1-6=0, so 1 is indeed a root. 

here i have given 3 method to factorise..

hope u like it..