Question

Verify that $3,-1,-\frac{1}{3}$ are the zeroes of the cubic polynomial

$p\left(x\right)=3x^3-5x^2-11x-3$ and then verify the relationship between the zeroes and the coefficients

Answer 1

We have,

$p\left(x\right)=3x^3-5x^2-11x-3$

And the zeros are

$3,-1,\frac{-1}{3}$

Verifying the zeros,

$x=3,p\left(3\right)=3(3{)}^3-5(3{)}^2-11\left(3\right)-3$

$=81-45-33-3$

$=0$

=$x-1,p(-1)=3(-1{)}^3-5(-1{)}^2-11(-1)-3$

$=-3-5+11-3$

=$0$

$x=\frac{-1}{3},$

$p\left(\frac{-1}{3}\right)=3(\frac{-1}{3}{)}^3-5(\frac{-1}{3}{)}^2-11\left(\frac{-1}{3}\right)-3$

$\frac{-1}{9}-\frac{5}{9}+\frac{11}{3}-3$

$\frac{-1-5-33-27}{9}$

$0$

Now verifying the relation between zeros and coefficients is:

for,

$p\left(x\right)=3x^3-5x^2-11x-3$

$a=3,b=-1,c=-11,d=-1$

and zeros $\alpha =3,\beta =-1\gamma =\frac{-1}{3}$

Now,

$\alpha +\beta +\gamma$

$=3+(-1)+\frac{-1}{3}$

$=\frac{9-3-1}{3}$

$\frac{5}{3}=\frac{-b}{a}$

$\alpha \beta +\beta \gamma +\gamma \alpha$

$=\left(3\right)(-1)+(-1)\left(\frac{-1}{3}\right)+\left(\frac{-1}{3}\right)\left(3\right)$

$\frac{-9+1-3}{3}$

$=\frac{-11}{3}=\frac{c}{a}$

$\alpha \beta \gamma$

$=\left(3\right)(-1)\left(\frac{-1}{3}\right)$

$1=\frac{d}{a}$

Thus the relation are verified.