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

Answer 1

The correct option is A $b-a+1$

Let $\alpha ,\beta$ be the other zeros of the given polynomial $x^3+a{x}^2+b{x}^2+c$

Sum of the zeros $=\frac{-\text{coefficient of}{x}^2}{\text{coefficient of}{x}^3}$

$\Rightarrow -1+\alpha +\beta =\frac{-a}{1}=-a$

$\Rightarrow \alpha +\beta =-a+1$ (i)

Again,

$(-1)\alpha +\alpha \beta +(-1)\beta =\frac{-\text{coefficient of}x}{\text{coefficient of}{x}^3}$

$\Rightarrow -\alpha +\alpha \beta -\beta =\frac{b}{1}$

$=\alpha \beta =b+\alpha +\beta$

$\alpha +\beta =-a+1$ , from (i))

$=b-a+1$