Question

If the product of two zeroes of the polynomial $p\left(x\right)=x^3-{6x}^2+11x-6$ is $3$, then the find its third zero.

Answer 1

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

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

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

Here $p\left(x\right)=x^3-6x^2+11x-6$ is of the form $a{x}^3+b{x}^2+cx+d$

$\therefore a=1,b=-6,c=11,d=-6$

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

$\Rightarrow \alpha \beta \gamma =6$

$\Rightarrow \gamma =\frac{6}{\alpha \beta }$

$\Rightarrow \gamma =\frac{6}{3}$ [from (1)]

$\Rightarrow \gamma =2$

Hence the third zero is 2.