Question

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

(a) $b-a+1$ (b) $b-a-1$ (c) $a-b+1$ (d) $a-b-1$

Answer 1

Let $p\left(x\right)=x^3+a{x}^2+bx+c$.
Now, −1 is a zero of the polynomial.
So, p(−1) = 0.
$$\begin{gathered} \Rightarrow {\left(-1\right)}^3+a{\left(-1\right)}^2+b\left(-1\right)+c=0 \\ \Rightarrow -1+a-b+c=0 \\ \Rightarrow a-b+c=1 \\ \Rightarrow c=1-a+b \end{gathered}$$
Now, if $\alpha ,\beta ,\gamma$ are the zeroes of the cubic polynomial $a{x}^3+b{x}^2+cx+d$, then product of zeroes is given by
$\alpha \beta \gamma =-\frac{d}{a}$
So, for the given polynomial, $p\left(x\right)=x^3+a{x}^2+bx+c$
$$\begin{gathered} \alpha \beta \left(-1\right)=\frac{-c}{1}=\frac{-\left(1-a+b\right)}{1} \\ \Rightarrow \alpha \beta =1-a+b \end{gathered}$$

Hence, the correct answer is option A.