Question

The product of zeros of the cubic polynomial $p\left( x \right)$ is:
(a) $\dfrac{\text{Coefficient of }{{x}^{2}}}{\text{Coefficient of }{{x}^{3}}}$
(b) $\dfrac{\text{Coefficient of }x}{\text{Coefficient of }{{x}^{3}}}$
(c) $-\left( \dfrac{\text{Constant Term}}{\text{Coefficient of }{{x}^{3}}} \right)$
(d) None of the above

Answer 1

Hint: We solve this problem by taking the examples of any 2 cubic polynomials. We take some zeros and find the equation of the cubic polynomial and then find the product of zeros we assumed and compare them with the cubic polynomial that is formed and check the options to find the correct answer.

Complete step by step solution:
We are asked to find the products of zeros of cubic polynomials.
Now, let us assume some cubic polynomials and find the product of zeros with respect to the equation.
Let use assume that the cubic polynomial $p\left( x \right)$ have zeros $1,2,3$ then we get the polynomial as,
$\Rightarrow p\left( x \right)=\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)$
Now, let us multiply the first two terms in the above polynomial then we get,
$\begin{align}
  & \Rightarrow p\left( x \right)=\left( {{x}^{2}}-x-2x+2 \right)\left( x-3 \right) \\
 & \Rightarrow p\left( x \right)=\left( {{x}^{2}}-3x+2 \right)\left( x-3 \right) \\
\end{align}$
Now, let us multiply the two terms in above equation then we get,
$\begin{align}
  & \Rightarrow p\left( x \right)={{x}^{3}}-3{{x}^{2}}-3{{x}^{2}}+9x+2x-6 \\
 & \Rightarrow p\left( x \right)={{x}^{3}}-6{{x}^{2}}+11x-6 \\
\end{align}$
Now, let us find the product of zeros of the above polynomial.
We know that the zeros of the above polynomial are $1,2,3$. By finding the product we get,
$\begin{align}
  & \Rightarrow p=1\times 2\times 3 \\
 & \Rightarrow p=6 \\
\end{align}$
Now, let us compare the above product with the options and the coefficients of polynomial then we get,
$\begin{align}
  & \Rightarrow p=-\left( \dfrac{-6}{1} \right) \\
 & \Rightarrow p=-\left( \dfrac{\text{Constant Term}}{\text{Coefficient of }{{x}^{3}}} \right) \\
\end{align}$
Now, let us verify this with another example.
Let use assume that the cubic polynomial $p\left( x \right)$ have zeros $0,2,3$ then we get the polynomial as,
$\Rightarrow p\left( x \right)=\left( x-0 \right)\left( x-2 \right)\left( x-3 \right)$
Now, let us multiply the first two terms in the above polynomial then we get,
$\Rightarrow p\left( x \right)=\left( {{x}^{2}}-2x \right)\left( x-3 \right)$
Now, let us multiply the two terms in above equation then we get,
$\begin{align}
  & \Rightarrow p\left( x \right)={{x}^{3}}-3{{x}^{2}}-2{{x}^{2}}+6x \\
 & \Rightarrow p\left( x \right)={{x}^{3}}-5{{x}^{2}}+6x \\
\end{align}$
Now, let us find the product of zeros of the above polynomial.
We know that the zeros of the above polynomial are $1,2,3$. By finding the product we get,
$\begin{align}
  & \Rightarrow p=0\times 2\times 3 \\
 & \Rightarrow p=0 \\
\end{align}$
Now, let us compare the above product with the options and the coefficients of polynomial then we get,
$\begin{align}
  & \Rightarrow p=-\left( \dfrac{0}{1} \right) \\
 & \Rightarrow p=-\left( \dfrac{\text{Constant Term}}{\text{Coefficient of }{{x}^{3}}} \right) \\
\end{align}$
Therefore, we can conclude that the product of zeros of the given cubic polynomial can be given as,
$\therefore $ Product of zeros is $-\left( \dfrac{\text{Constant Term}}{\text{Coefficient of }{{x}^{3}}} \right)$

So, the correct answer is “Option c”.

Note: We can directly give the answer by using the generalisation.
We have the condition that if there is a ${{n}^{th}}$ degree polynomial $p\left( x \right)$ in the form,
$\Rightarrow p\left( x \right)={{a}_{0}}{{x}^{n}}+{{a}_{1}}{{x}^{n-1}}+{{a}_{2}}{{x}^{n-2}}+......+{{a}_{n-2}}{{x}^{2}}+{{a}_{n-1}}x+{{a}_{n}}$
Also if $'{{S}_{n}}'$ denotes the sum of product of roots taken $'n'$ at a time that is,
 $'{{S}_{1}}'$ is the sum of the product of roots taken 1 at a time.
 $'{{S}_{2}}'$ is the sum of product of roots taken 2 at a time and so on then,
${{S}_{n}}={{\left( -1 \right)}^{n}}\left( \dfrac{{{a}_{n}}}{{{a}_{0}}} \right)$
Now, let us consider the cubic polynomial which will be in the form,
$\Rightarrow p\left( x \right)={{a}_{0}}{{x}^{3}}+{{a}_{1}}{{x}^{2}}+{{a}_{2}}x+{{a}_{3}}$
Now, we know that the product of roots is obtained by taking all 3 roots at a time so that the product of roots can be represented as ${{S}_{3}}$
So, by using the above formula we get the value of ${{S}_{3}}$ as,
$\begin{align}
  & \Rightarrow {{S}_{3}}={{\left( -1 \right)}^{3}}\left( \dfrac{{{a}_{3}}}{{{a}_{0}}} \right) \\
 & \Rightarrow {{S}_{3}}=-\left( \dfrac{{{a}_{3}}}{{{a}_{0}}} \right) \\
\end{align}$
Now, by comparing the above equation with respect to the coefficients of polynomial then we get,
$\Rightarrow {{S}_{3}}=-\left( \dfrac{\text{Constant Term}}{\text{Coefficient of }{{x}^{3}}} \right)$
Therefore, we can conclude that the product of zeros of the given cubic polynomial can be given as,
$\therefore $ Product of zeros is $-\left( \dfrac{\text{Constant Term}}{\text{Coefficient of }{{x}^{3}}} \right)$
So, option (c) is the correct answer.