Question

The product of two zeros of polynomial. $P\left(x\right)=x^3-3x^3-6x+8$ is $(-2)$. Find all the zeros of the polynomial.

Answer 1

$\begin{array}{c}P\left(x\right)=x^3-3x^2-6x+8=0\\ letthreezerioesare\lambda ,\beta and\gamma \\ fromquestion\lambda \beta =-2\\ \Rightarrow \beta =\frac{-2}{\lambda }\\ productofzeroes,\lambda \beta \gamma =\frac{-d}{a}\\ -2r=\frac{-8}{1}\\ \gamma =\frac{-8}{-2}=4\\ sumofzeroes=\lambda +\beta +\gamma =\frac{-b}{a}\\ \lambda +\frac{-2}{\lambda }+4=\frac{-\left(-3\right)}{1}\\ {\lambda }^2-2+4\lambda =3\lambda \\ {\lambda }^2+\lambda -2=0\\ {\lambda }^2+2\lambda -\lambda -2=0\\ \lambda \left(\lambda +2\right)-1\left(\lambda +2\right)=0\\ \left(\lambda +2\right)\left(\lambda -1\right)=0\\ \therefore \lambda =-2or-1\\ \therefore \beta =\frac{-2}{-2}=1or\frac{-2}{-1}=2\end{array}$