Question

Given the zeroes of a cubic polynomial $p\left(x\right)=3x^3-10x^2-27x+10$ are 5, -2 and $\frac{1}{3}$. Verify the relation between its zeros and coefficients.

Answer 1

$p\left(x\right)=3x^3–10x^2–27x+10$

Therefore, $\alpha =5,\beta =-2,\gamma =\frac{1}{3}$

Comparing the given polynomial with

$p\left(x\right)=a{x}^3+b{x}^2+cx+d$

We get $a=3,b=-10,c=-27andd=10$

Now, $(\alpha +\beta +\gamma )=(5-2+\frac{1}{3})=\frac{10}{3}=\frac{-b}{a}$

$(\alpha \beta +\beta \gamma +\gamma \alpha )=[5\times (-2)+(-2)\times \frac{1}{3}+\frac{1}{3}\times 5)]=(-10-\frac{2}{3}+\frac{5}{3})=\frac{-30-2+5}{3}=\frac{-27}{3}=\frac{c}{a}$

and $\alpha \beta \gamma =[5\times (-2)\times \frac{1}{3}]=-\frac{10}{3}=-\frac{d}{a}$