Question

Verify that $3,-1,\dfrac{-1}{3}$ are the zeroes of the cubic polynomial $p\left( x \right)=3{{x}^{3}}-5{{x}^{2}}-11x-3$, and then verify the relationship between the zeroes and the coefficients.

Answer 1

Hint: Substitute the given values in the equation of the cubic polynomial and check if they satisfy the given equation or not. Use the fact that if $\alpha ,\beta ,\gamma $ are roots of the cubic equation of the form $a{{x}^{3}}+b{{x}^{2}}+cx+d$, then we have $\alpha +\beta +\gamma =\dfrac{-b}{a}$, $\alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{c}{a}$ and $\alpha \beta \gamma =\dfrac{-d}{a}$.

Complete step-by-step solution -
We have to verify that $3,-1,\dfrac{-1}{3}$ are the zeroes of the cubic polynomial $p\left( x \right)=3{{x}^{3}}-5{{x}^{2}}-11x-3$ and then verify the relationship between the zeroes and the coefficients.
To verify that $3,-1,\dfrac{-1}{3}$ are the zeroes of the cubic polynomial $p\left( x \right)=3{{x}^{3}}-5{{x}^{2}}-11x-3$, we will substitute the values in the polynomial equation and check if they satisfy the equation or not.
Substituting $x=3$ in the cubic equation $p\left( x \right)=3{{x}^{3}}-5{{x}^{2}}-11x-3$, we have $p\left( 3 \right)=3{{\left( 3 \right)}^{3}}-5{{\left( 3 \right)}^{2}}-11\left( 3 \right)-3=81-45-33-3=81-81=0$.
Substituting $x=-1$ in the cubic equation $p\left( x \right)=3{{x}^{3}}-5{{x}^{2}}-11x-3$, we have $p\left( -1 \right)=3{{\left( -1 \right)}^{3}}-5{{\left( -1 \right)}^{2}}-11\left( -1 \right)-3=-3-5+11-3=11-11=0$.
Substituting $x=\dfrac{-1}{3}$ in the cubic equation $p\left( x \right)=3{{x}^{3}}-5{{x}^{2}}-11x-3$, we have $p\left( \dfrac{-1}{3} \right)=3{{\left( \dfrac{-1}{3} \right)}^{3}}-5{{\left( \dfrac{-1}{3} \right)}^{2}}-11\left( \dfrac{-1}{3} \right)-3=\dfrac{-1}{9}-\dfrac{5}{9}+\dfrac{11}{3}-3=\dfrac{-1-5+33-27}{9}=\dfrac{32-32}{9}=0$.
As $3,-1,\dfrac{-1}{3}$ satisfy the cubic polynomial $p\left( x \right)=3{{x}^{3}}-5{{x}^{2}}-11x-3$, they are the roots of the given equation.
We will now verify the relation between the roots and the coefficients of the cubic polynomial. We know that if $\alpha ,\beta ,\gamma $ are roots of the cubic equation of the form $a{{x}^{3}}+b{{x}^{2}}+cx+d$, then we have $\alpha +\beta +\gamma =\dfrac{-b}{a}$, $\alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{c}{a}$ and $\alpha \beta \gamma =\dfrac{-d}{a}$.
Substituting $\alpha =3,\beta =-1,\gamma =\dfrac{-1}{3},a=3,b=-5,c=-11,d=-3$ in the above expression, we have
$\alpha +\beta +\gamma =\dfrac{-b}{a}$
$\Rightarrow 3+\left( -1 \right)+\left( \dfrac{-1}{3} \right)=\dfrac{-\left( -5 \right)}{3}$
$\Rightarrow 2-\dfrac{1}{3}=\dfrac{5}{3}$
$\Rightarrow \dfrac{6-1}{3}=\dfrac{5}{3}$
$\Rightarrow \dfrac{5}{3}=\dfrac{5}{3} $.
Similarly, we have
$\alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{c}{a}$
$\Rightarrow 3\left( -1 \right)+\left( -1 \right)\left( \dfrac{-1}{3} \right)+3\left( \dfrac{-1}{3} \right)=\dfrac{-11}{3}$
$\Rightarrow -3-1+\dfrac{1}{3}=\dfrac{-11}{3}$
$\Rightarrow \dfrac{1}{3}-4=\dfrac{-11}{3}$
$\Rightarrow \dfrac{1-12}{3}=\dfrac{-11}{3}$
$\Rightarrow \dfrac{-11}{3}=\dfrac{-11}{3}$
and
$\alpha \beta \gamma =\dfrac{-d}{a}$.
$\Rightarrow 3\left( -1 \right)\left( \dfrac{-1}{3}\right)=\dfrac{-\left( -3 \right)}{3}$
$\Rightarrow 1=1 $.
Hence, we have verified that $3,-1,\dfrac{-1}{3}$ are the zeroes of the cubic polynomial $p\left( x \right)=3{{x}^{3}}-5{{x}^{2}}-11x-3$ and we have also verified the relation between the roots and the coefficients of the cubic equation.

Note: We can also solve this question by factorizing the given cubic polynomial by splitting the middle term and then calculating the roots of the cubic equation to check if they match with the roots given in the question or not.