Question

If one of the zeros of the cubic polynomial \[{x^3} + a{x^2} + bx + c\] is −1, then the product of the other two zeros is
\[
  A.{\text{ }}b - a + 1 \\
  B.{\text{ }}b - a - 1 \\
  C.{\text{ }}a - b + 1 \\
  D.{\text{ }}a - b - 1 \\
 \]

Answer 1

Hint: In order to find out the product of other two zeros, we will use two properties of cubic polynomial, first one is that sum of zeros of cubic polynomial is equal to the ratio of coefficient of \[{x^2}\] to the coefficient of \[{x^3}\] and multiplied with -1, and second one is that sum of three pair of product of two zeros is equal to the ratio of coefficient of $x$ to the coefficient of \[{x^3}\] and multiplied with -1.

Complete step-by-step answer:
Given cubic polynomial is \[{x^3} + a{x^2} + bx + c\] and one zero of this polynomial is -1
Let \[\alpha ,\beta \] be the other zeros of the given polynomial \[{x^3} + a{x^2} + bx + c\]
We know that:
Sum of zeros of cubic polynomial is equal to the ratio of coefficient of \[{x^2}\] to the coefficient of \[{x^3}\] and multiplied with -1.
So, sum of the zeros \[ = \dfrac{{ - {\text{coefficient of }}{x^3}}}{{{\text{coefficient of }}{x^2}}}\]
So, we have:
$
   \Rightarrow - 1 + \alpha + \beta = \dfrac{{ - a}}{1} = - a \\
   \Rightarrow \alpha + \beta = - a + 1.............(1) \\
 $
We also know that:
Sum of three pairs of products of two zeros is equal to the ratio of coefficient of $x$ to the coefficient of \[{x^3}\] and multiplied with -1.
So, we have:
\[
   \Rightarrow \left( { - 1} \right)\alpha + \alpha \beta + \left( { - 1} \right)\beta = \dfrac{{ - {\text{coefficient of }}x}}{{{\text{coefficient of }}{x^3}}} \\
   \Rightarrow - \alpha + \alpha \beta - \beta = \dfrac{b}{1} \\
   \Rightarrow \alpha \beta = b + \alpha + \beta ..........(2) \\
 \]
Now let us substitute the value of equation (1) into equation (2)
\[
   \Rightarrow \alpha \beta = b + \left( {\alpha + \beta } \right) \\
   \Rightarrow \alpha \beta = b + \left( { - a + 1} \right){\text{ }}\left[ {{\text{from equation (1)}}} \right] \\
   \Rightarrow \alpha \beta = b - a + 1 \\
 \]
Hence, the product of other roots of the equation is \[b - a + 1\]
So, the correct answer is option A.

Note- This problem can also be solved by finding the other two roots of the cubic polynomial and hence finding the product of the other two roots. But in order to solve such problems where directly the roots are not asked and only some functions of the roots are asked, the different relations between the roots are important for solving such problems. These relations must be remembered by the students.