Question

How do you solve \[{{x}^{3}}+3{{x}^{2}}-x-3=0\]?

Answer 1

Hint: A cubic polynomial has three roots. We can solve the above equation by using the factorisation method. factorisation or factoring is defined as the breaking of an entity into a product of their factors which when multiplied together give the original number. In the factorisation method, we reduce any algebraic or cubic equation into its simpler form, where the equations are represented as the product of their factors instead of expanding the brackets.

Complete step by step answer:
As per the given question, we have to find the factors of the given cubic polynomial and solve the factors to find the roots of the equation. And the given polynomial to be factored is \[{{x}^{3}}+3{{x}^{2}}-x-3=0\].
In the given cubic polynomial \[{{x}^{3}}+3{{x}^{2}}-x-3=0\], \[{{x}^{2}}\] is common in the first 2 terms. Since \[{{x}^{3}}\]can be written as \[{{x}^{2}}\times x\]. -1 is common in the last 2 terms. Now, we take the terms common in the first 2 terms and last 2 terms.
\[\begin{align}
  & \Rightarrow {{x}^{3}}+3{{x}^{2}}-x-3=0 \\
 & \Rightarrow {{x}^{2}}(x+3)-1(x+3)=0 \\
\end{align}\]
Here, we have \[(x+3)\] in common then
\[\Rightarrow ({{x}^{2}}-1)(x+3)=0\]
Now we equate the equation, then we get
\[\begin{align}
  & \Rightarrow ({{x}^{2}}-1)=0,(x+3)=0 \\
 & \Rightarrow {{x}^{2}}=1,x=-3 \\
 & \Rightarrow x=\pm 1,x=-3 \\
 & \Rightarrow x=1,-1,-3 \\
\end{align}\]
\[\therefore x=1,-1,-3\] are the roots of the given cubic equation \[{{x}^{3}}+3{{x}^{2}}-x-3=0\].

Note: While solving such types of problems, there is a scope of making mistakes when we try to make a factor common. Calculation mistakes should be avoided to get the correct answer. In some cubic equations, the equation cannot be completely factored. In those cases we find a root using trial and error method and solve the equation using a long division method then we get a quadratic equation. we solve the quadratic equation and find the other two roots of the equation.