Question

If one of the zeros of the cubic polynomial x³ + ax² + bx + c is −1, then the product of the other two zeros is

(a) a − b − 1
(b) b − a − 1
(c) 1 − a + b
(d) 1 + a − b

Answer 1

$$\begin{gathered} \left(\text{c}\right)1-a+b \\ \text{Since}-1\text{is a zero of}{x}^3+a{x}^2+bx+c,\text{we have:} \\ {(-1)}^3+a\times {(-1)}^2+b\times (-1)+c=0 \\ =>a-b+c-1=0 \\ =>c=1-a+b \\ \text{Also, product of all zeroes is given by} \\ \alpha \beta \times (-1)=-c \\ =>\alpha \beta =c \\ =>\alpha \beta =1-a+b \end{gathered}$$