Question

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

Answer 1

Hint: Hint:-Here in this question we have to find all the roots of the cubic polynomial. For this we will apply the sum of roots and product of roots formula. The general cubic polynomial equation is given by \[A{x^3} + B{x^2} + Cx + D = 0\] and if $\alpha ,\beta , \gamma $ are three roots.
The sum of roots $\alpha + \beta + \gamma = \dfrac{{ - B}}{A}$
The product of roots $\alpha \beta \gamma = \dfrac{{ - D}}{A}$
Sum of product of two roots $\alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{C}{A}$

Complete step-by-step answer:
Given cubic polynomial is \[{x^3} + a{x^2} + bx + c\] and one of the roots is -1. Let $\alpha = - 1$ so when comparing with the general equation \[A{x^3} + B{x^2} + Cx + D = 0\] we get that coefficients A=1, B=a, C=b and D=c.
Now we will apply the sum of roots formula.
The sum of roots $ - 1 + \beta + \gamma = \dfrac{{ - a}}{1}$
$ \Rightarrow \beta + \gamma = - a + 1$ Equation (1)
Now we will apply the product of roots formula.
The product of roots $\alpha \beta \gamma = \dfrac{{ - D}}{A}$
$ \Rightarrow - \beta \gamma = \dfrac{{ - c}}{1}$
$ \Rightarrow \beta \gamma = c$ Equation (2)
Now we will apply the sum of products of two roots formula.
Sum of product of two roots $\alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{C}{A}$
$ \Rightarrow - \beta + \beta \gamma - \gamma = \dfrac{b}{1}$
Now we will use equation (2) and put value of $\beta \gamma = c$
$ \Rightarrow - \beta + c - \gamma = b$
\[ \Rightarrow c - b = \beta + \gamma \]
Now we will use equation (1) and put value of $\beta + \gamma = - a + 1$
\[ \Rightarrow c - b = - a + 1\]
\[ \Rightarrow c = b - a + 1\]
Now we will put this value of c in equation 2 and we will get,
$ \Rightarrow \beta \gamma = b - a + 1$
As $\beta ,\gamma $ are the other two roots of the given cubic polynomial therefore their product is $b-a+1$.

Option A is the correct answer.

Note: Students may likely make mistakes by finding all the two roots and then multiplying them but that process will be too time confusing instead this above method is efficient as direct product formulas have been used so students must have a good understanding of these formulas.