Question

If the product of two zeroes of the polynomial $p\left(x\right)={2x}^3+{6x}^2-4x+9$ is $3$, then the find its third zero.

Answer 1

Let $\alpha ,\beta ,\gamma$ zeros of the polynomial p(x).

It is given that the product of two zeros of p(x) is 3

ie $\alpha \beta =3$ ________ (1)

Here $p\left(x\right)=2x^3-6x^2-4x-9$ is of the form $a{x}^3+b{x}^2+cx+d$

$\therefore a=2,b=6,c=-4,d=9$

Then $\alpha \beta \gamma =\frac{-d}{a}$

$\Rightarrow \alpha \beta \gamma =\frac{-9}{2}$

$\Rightarrow \gamma =\frac{-9}{2}.\frac{1}{\alpha \beta }$

$\Rightarrow \gamma =\frac{-9}{2}.\frac{1}{3}$ [from (1)]

$\Rightarrow \gamma =\frac{-3}{2}$

Hence the third zero is= -3/2.