Question

Factorise:-
          ${{x}^{3}}-3{{x}^{2}}-9x-5$

Answer 1

Hint: Whenever, there is a polynomial, whose degree is greater than 2, then we have to identify a root by hit and trial method. After we get its one root, the remaining two can also be identified easily.

Complete step by step answer:
Hence, to factorize the above equation, the first step that we have to take is to find its one root using the hit and trial method.
Hit and trial method is used to find the factors or roots of any polynomial of degree more than two. In this method generally, we try to put some different-different values of the numbers in the given polynomial until the polynomial satisfies zero of that polynomial.
But, in general, one should always start putting the integers, whose range is from -2 to +2 in the order 0,+1,-1,+2,-2. In most of the cases, you will find a root among these.
So, let us start accordingly.
For $x=0,$ the value of the given polynomial is:
$\begin{align}
  & {{x}^{3}}-3{{x}^{2}}-9x-5 \\
 & \Rightarrow {{0}^{3}}-3\times {{0}^{2}}-9\times 0-5=-5 \\
\end{align}$
But, $-5\ne 0$
So, (x) is not a factor of the above polynomial.
For $x=1,$ the value of the given polynomial is:
$\begin{align}
  & {{x}^{3}}-3{{x}^{2}}-9x-5 \\
 & \Rightarrow {{1}^{3}}-3\times {{1}^{2}}-9\times 1-5=-16 \\
\end{align}$
But, $-16\ne 0$
So, (x-1) is not a factor of the above polynomial.
For $x=-1,$ the value of the given polynomial is:
$\begin{align}
  & {{x}^{3}}-3{{x}^{2}}-9x-5 \\
 & \Rightarrow {{(-1)}^{3}}-3\times {{(-1)}^{2}}-9\times (-1)-5=0 \\
\end{align}$
$\Rightarrow x=-1$ is a zero of the given polynomial.
Hence, (x+1) is a factor of the above polynomial.
Now, when the polynomial ${{x}^{3}}-3{{x}^{2}}-9x-5$ is divided by (x+1), the quotient is ${{x}^{2}}-4x-5$ and the remainder is 0.
We know that: Dividend = (Divisor x Quotient) + Remainder
\[\begin{align}
  & \Rightarrow {{x}^{3}}-3{{x}^{2}}-9x-5=(x+1)\times ({{x}^{2}}-4x-5)+0 \\
 & \Rightarrow {{x}^{3}}-3{{x}^{2}}-9x-5=(x+1)\times ({{x}^{2}}-4x-5) \\
 & \\
\end{align}\]
Now, the polynomial ${{x}^{2}}-4x-5$ can further be factorized by the method of splitting the middle-term.
Therefore, the polynomial ${{x}^{2}}-4x-5$ can be written as $\begin{align}
  & {{x}^{2}}-4x-5={{x}^{2}}-5x+x-5 \\
 & \Rightarrow x(x-5)+1(x-5) \\
\end{align}$
Now, taking (x-5) as common, we can write it as: $\Rightarrow x(x-5)+1(x-5)=(x-5)(x+1)$
So, ${{x}^{2}}-4x-5=(x-5)(x+1)$
Now, the polynomial given in the question can be written in a way:
\[\begin{align}
  & \Rightarrow {{x}^{3}}-3{{x}^{2}}-9x-5=(x+1)\times ({{x}^{2}}-4x-5) \\
 & \Rightarrow {{x}^{3}}-3{{x}^{2}}-9x-5=(x+1)(x-5)(x+1)=(x-5){{(x+1)}^{2}} \\
\end{align}\]
Hence, the polynomial is factored.

Note: While doing these types of factorization questions, we can also simply use the method of splitting them. That will make our question short but it will involve mental calculations.