Question

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

Answer 1

$p\left(x\right)=3x^3–10x^2–27x+10$

Therefore, $p\left(5\right)=3(5{)}^3–10(5{)}^2–27\left(5\right)+10$

$=375–250–135+10=0$

$p(-2)=3(-2{)}^3–10(-2{)}^2–27(-2)+10$

$=-24–40+54+10=0$

$p\left(1/3\right)=3(1/3{)}^3–10(1/3{)}^2–27\left(1/3\right)+10$

$=1/9-10/9-9+10=0$

Therefore, Then, 5, -2, $\frac{1}{3}$ zeros of

$p\left(x\right)=3x^3–10x^2–27x+10$

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