Question

How to solve the \[A\] method for \[ a{{x}^{3}}+b{{x}^{2}}+cx+d=0 \] ?

Answer 1

Hint: In this type of question you need to first know the basic definition of cubic equation and what it means, then you can basically try to find one of its root and make it in the form of quadratic equation

Complete step-by-step solution:
In algebra, a cubic equation in one variable is an equation of the form \[a{{x}^{3}}+b{{x}^{2}}+cx+d=0\] in which \[a\] is nonzero.
The solutions of this equation are called roots of the cubic function defined by the left-hand side of the equation. If all of the coefficients \[a,b,c\] and \[d\] of the cubic equation are real numbers, then it has at least one real root (this is true for all odd-degree polynomial functions).
Now we are going to see how we can solve such cubic equations.
The traditional way of solving a cubic equation is to reduce it to a quadratic equation and then solve it either by factoring or quadratic formula.
Like a quadratic equation has two real roots, a cubic equation may have possibly three real roots. But unlike a quadratic equation, which may have no real solution, a cubic equation has at least one real root.
The other two roots might be real or imaginary.
Whenever you are given a cubic equation or any equation, you always have to arrange it in a standard form first.
For example, if you are given something like this, \[3{{x}^{2}}+x-3=\dfrac{2}{x}\] , you will re-arrange into the standard form and write it like, \[3{{x}^{3}}+{{x}^{2}}-3x-2=0\] . Then you can solve this by any suitable method.
Let’s see a few examples below for better understanding:
Suppose we have to find the roots of equation:
\[{{x}^{3}}-6{{x}^{2}}+11x-6=0\]
So we will try to first find its one real root by hit and trial since we know that the cubic equation holds at least one real root.
So we have,
\[(x-1)\] is one of the factors since, by putting \[x=1\] the equation is satisfied.
So now by dividing \[{{x}^{3}}-6{{x}^{2}}+11x-6\] by \[(x-1)\] ,
\[\Rightarrow \left( x-\text{ }\text{ }1 \right)\left( {{x}^{2}}-\text{ }5x\text{ }+\text{ }6 \right)=0\]
Now we have reduced our equation to quadratic equation, now we can easily factorise it or find other roots.
So we have,
\[\Rightarrow \left( x-\text{ }\text{ }1 \right)\left( x-\text{ }\text{ }2 \right)\left( x-\text{ }\text{ }3 \right)=0\]
Thus the above cubic equation solutions are \[x=1,x=2\] and \[x=3\] .
Therefore this is one of the ways or methods we can solve \[a{{x}^{3}}+b{{x}^{2}}+cx+d=0\] such an equation.

Note: The first general solution of the cubic equation was found by Scipione Del Ferro at the beginning of the 16th century and rediscovered by Niccolò Tartaglia several years later. The solution was published by Gerolamo Cardano in his Ars magna (Ars Magna or the Rules).