Question

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

Answer 1

We have,

$x^3+{ax}^2+bx+c=0$

Now say $\alpha$ and $\beta$ are the two roots while ${}^{\prime }-1^{\prime }$ is also root of the above abic polynomial

Now,

Sum of roots $=\alpha +\beta -1=-\frac{a}{1}=-a⟶\left(1\right)$

products of roots taken two at a time $=\alpha \beta -\alpha -\beta =b⟶\left(2\right)$

products of roots $=\alpha \beta (-1)=-c⟶\left(3\right)$

$\therefore$ from $\left(1\right)$ we have,

$\alpha +\beta =1-a⟶\left(4\right)$

from $\left(2\right)$

$\alpha \beta -(\alpha +\beta )=b$

$\Rightarrow \alpha \beta -(1-a)=b$ $......$ $($from $\left(4\right))$

$\Rightarrow \alpha \beta =b+1-a⟶\left(5\right)$

from $\left(3\right)$

$+\alpha \beta =+c$

$\alpha \beta =c⟶\left(6\right)$

$\therefore$ comparing $\left(5\right)$ & $\left(6\right)$ we get

$b+1-a=c\Rightarrow a-b+c=1⟶\left(7\right)$

Now since ${}^{\prime }-1^{\prime }$ is a root of the poly nomial

$\therefore$ ${(-1)}^3+a{(-1)}^2+b(-1)+c=0$

$\Rightarrow -1+a-b+c=0$

$\Rightarrow a-b+c=1$

$\therefore$ the product of other two zeroes is given by $\left(6\right)$ or $\left(5\right)$

i.e. $\alpha \beta =c=b-a+1$