Question

Question 6
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 zeros is
(A) b – a + 1
(B) b – a – 1
(C) a – b + 1
(D) a – b –1

Answer 1

Let $p\left(x\right)=x^3+a{x}^2+bx+c$
Let $\alpha \beta$ and $\gamma$ be the zeroes of the given cubic polynomial p (x)
$\therefore \alpha =-1$ [given]
and p( - 1) = 0
$\Rightarrow$ $(-1{)}^3+a(-1{)}^2+b(-1)+c=0$
$\Rightarrow$ $-1+a-b+c=0$
$\Rightarrow$ $c=1–a+b$
We know that,
Product of the zeroes = $(-1{)}^3\frac{constantterm}{coefficientof{x}^3}=-\frac{c}{1}$
$\alpha \beta \gamma =-c$
$\Rightarrow (-1)\beta \gamma =-c[\because \alpha =-1]$
$\Rightarrow \beta \gamma =c$
$\Rightarrow \beta \gamma =1-a+b[fromEq.(i\left)\right]$
Hence, product of the other two roots is 1 – a + b