Question 6

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

Solution:

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 $p\left(x\right)$.
$\therefore \alpha =-1$” “[given]
and $p(-1)=0$
$\Rightarrow {(-1)}^3+a{(-1)}^2+b(-1)+c=0$
$\Rightarrow -1+a-b+c=0$
$\Rightarrow c=1-a+b$ …(i)
We know that,
Product of all zeroes $={(-1)}^{\mathrm{3[}}\frac{\text{Constant term/}}{\text{Coefficient of}}=-\frac{c}{1}$
$\alpha \beta \gamma =-c$
$\to (-1)\beta \gamma =-c[\because \alpha =-1]$
$\Rightarrow \mathrm{\beta \gamma }=c$
$\Rightarrow \beta \gamma =1-a+b$ ” “[from Eq.(i)]
Hence, product the other two roots is $1-a+b$.