Question

If one of the zeroes of the cubic polynomial $x^3+a{x}^2+bx+c$ is $-1$, then the product of the other two zeroes is

• A. $b-a+1$
• B. $b-a-1$
• C. $a-b+1$
• D. $a-b-1$

Answer 1

The correct option is A

$b-a+1$

Solution:

Finding the relation between zeroes and the coefficient:

Given: One of the zeroes of the cubic polynomial $x^3+a{x}^2+bx+c$ is $-1$.

Let the zeroes are $p,qandr=-1.$

Sum of the zeroes: $p+q+r=-a\left[\because sumofzeroes=\frac{-coefficientof{x}^2}{coefficientof{x}^3}\right]$

$$\begin{gathered} \Rightarrow p+q-1=-a\left[\because r=-1\right] \\ \Rightarrow p+q=1-a...\left(1\right) \end{gathered}$$

Sum of product of zeroes: $pq+qr+pr=b\left[\because sumofproductofzeroestakentwoatatime=\frac{coefficientofx}{coefficientof{x}^3}\right]$

$$\begin{gathered} \Rightarrow pq-q-p=b\left[\because r=-1\right] \\ \Rightarrow pq=b+p+q \\ \Rightarrow pq=b+1-a\left[fromeqn\left(1\right),p+q=1-a\right] \\ \mathbf{\Rightarrow }\mathit{p}\mathit{q}\mathbf{=}\mathit{b}\mathbf{-}\mathit{a}\mathbf{+}\mathbf{1} \end{gathered}$$

Final answer: Hence, the correct is option (A)