Question

Factorize \[{{x}^{3}}-\ 3{{x}^{2}}\ -\ 9x\ -5\] :-
a) \[{{\left( x\ +\ 1 \right)}^{2}}\ \left( x\ -\ 5 \right)\],
b) \[{{\left( x\ -\ 1 \right)}^{2}}\ \left( x\ -\ 5 \right)\],
c) \[\left( x\text{ }+\text{ }1 \right)\text{ }\left( x\text{ }+\text{ }5 \right)\text{ }\left( x\text{ }\text{ }1 \right)\],
d) None of these.

Answer 1

Hint: We start solving the problem assigning a variable for the given polynomial. We can see that the sum of the coefficients of odd powers of x is equal to the sum of the coefficients of even powers of x which leads to our first factor $\left( x+1 \right)$. We then divide the polynomial ${{x}^{3}}-3{{x}^{2}}-9x-5$ with $\left( x+1 \right)$ to get the quotient which contains multiplication of other factors. We then factorize the quotient and find the remaining two factors.

Complete step by step answer:
According to the problem, we have to find the factorization of the polynomial ${{x}^{3}}-3{{x}^{2}}-9x-5$.
Let us assume the given polynomial as $f\left( x \right)$.
So, we have $f\left( x \right)={{x}^{3}}-3{{x}^{2}}-9x-5$---(1).
Let us find the sum of the coefficients of odd powers of x.
We have a sum of odd powers of x = $1-9=-8$ ---(2).
Let us find the sum of the coefficients of even powers of x.
We have the sum of even powers of x = $-3-5=-8$ ---(3).
From (2) and (3), we can see that the sum of coefficients of odd powers of x is equal to the sum of even coefficients of x.
We know that if the sum of coefficients of odd powers of x is equal to the sum of even coefficients of x for a given polynomial then $\left( x+1 \right)$ will be a factor of that polynomial.
Let us divide the polynomial $f\left( x \right)$ with $\left( x+1 \right)$.
\[\begin{align}
  & \left. x+1 \right){{x}^{3}}-3{{x}^{2}}-9x-5\left( {{x}^{2}} \right.-4x-5 \\
 & \underline{\text{ }{{x}^{3}}+{{x}^{2}}\text{ }} \\
 & \text{ }0{{x}^{3}}-4{{x}^{2}}-9x-5 \\
 & \underline{\text{ }0{{x}^{3}}-4{{x}^{2}}-4x\text{ }} \\
 & \text{ }0{{x}^{3}}+0{{x}^{2}}-5x-5 \\
 & \underline{\text{ }0{{x}^{3}}+0{{x}^{2}}-5x-5\text{ }} \\
 & \text{ }0{{x}^{3}}+0{{x}^{2}}+0x+0 \\
\end{align}\].
So, we can write $f\left( x \right)$ as\[f\left( x \right)=\left( x+1 \right)\times \left( {{x}^{2}}-4x-5 \right)\].
Now we factorize ${{x}^{2}}-4x-5$.
\[f\left( x \right)=\left( x+1 \right)\times \left( {{x}^{2}}-5x+x-5 \right)\].
\[\Rightarrow f\left( x \right)=\left( x+1 \right)\times \left( x\left( x-5 \right)+1\left( x-5 \right) \right)\].
\[\Rightarrow f\left( x \right)=\left( x+1 \right)\times \left( \left( x+1 \right)\left( x-5 \right) \right)\].
\[\Rightarrow f\left( x \right)=\left( x+1 \right)\left( x+1 \right)\left( x-5 \right)\].
\[\Rightarrow f\left( x \right)={{\left( x+1 \right)}^{2}}\left( x-5 \right)\].
We have found the factorization of ${{x}^{3}}-3{{x}^{2}}-9x-5$ as \[{{\left( x+1 \right)}^{2}}\left( x-5 \right)\].

So, the correct answer is “Option a”.

Note: Alternatively, we can also solve the problem as follows:
$f\left( x \right)={{x}^{3}}-3{{x}^{2}}-9x-5$.
$\Rightarrow f\left( x \right)={{x}^{3}}+{{x}^{2}}-4{{x}^{2}}-4x-5x-5$.
$\Rightarrow f\left( x \right)={{x}^{2}}\left( x+1 \right)-4x\left( x+1 \right)-5\left( x+1 \right)$.
\[\Rightarrow f\left( x \right)=\left( x+1 \right)\left( {{x}^{2}}-4x-5 \right)\].
\[\Rightarrow f\left( x \right)=\left( x+1 \right)\times \left( {{x}^{2}}-5x+x-5 \right)\].
\[\Rightarrow f\left( x \right)=\left( x+1 \right)\times \left( x\left( x-5 \right)+1\left( x-5 \right) \right)\].
\[\Rightarrow f\left( x \right)=\left( x+1 \right)\times \left( \left( x+1 \right)\left( x-5 \right) \right)\].
\[\Rightarrow f\left( x \right)=\left( x+1 \right)\left( x+1 \right)\left( x-5 \right)\].
\[\Rightarrow f\left( x \right)={{\left( x+1 \right)}^{2}}\left( x-5 \right)\].