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 coefficient.

Answer 1

Hint: In this question, we need to verify the given zeroes of the polynomial p(x) and also the relationship between the zeroes and the coefficient. For verifying zero, we will put the value of x as given zeroes one by one in p(x) and prove it to be equal to zero. For verifying the relationship between zeroes and coefficient, we will verify following relation:
For zeroes $ \alpha ,\beta ,\gamma $ of a polynomial $ p\left( x \right)=a{{x}^{3}}+b{{x}^{2}}+cx+d $ .
\[\begin{align}
  & \text{Sum}\left( \alpha +\beta +\gamma \right)=\dfrac{-b}{a} \\
 & \text{Product}\left( \alpha \beta \gamma \right)=\dfrac{-d}{a} \\
 & \alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{c}{a} \\
\end{align}\]

Complete step by step answer:
Here we are given the polynomial p(x) as $ 3{{x}^{3}}-5{{x}^{2}}-11x-3 $ .
Zeroes of the polynomial are given as 3, -1, $ \dfrac{-1}{3} $ .
We need to verify them first. For this, let us put their values in place of x in p(x).
Putting x = 3 in p(x) we get,
 $ p\left( 3 \right)=3{{\left( 3 \right)}^{3}}-5{{\left( 3 \right)}^{2}}-11\left( 3 \right)-3 $ .
We know that $ {{\left( 3 \right)}^{3}}=27\text{ and }{{\left( 3 \right)}^{2}}=9 $ so we get,
\[\begin{align}
  & p\left( 3 \right)=3\left( 27 \right)-5\left( 9 \right)-11\left( 3 \right)-3 \\
 & \Rightarrow p\left( 3 \right)=81-45-33-3 \\
 & \Rightarrow p\left( 3 \right)=0 \\
\end{align}\]
Value of x as 3 gives p(x) as zero, therefore 3 is a zero of the polynomial.
Putting x = -1 in polynomial, we get,
\[\begin{align}
  & p\left( -1 \right)=3{{\left( -1 \right)}^{3}}-5{{\left( -1 \right)}^{2}}-11\left( -1 \right)-3 \\
 & \Rightarrow p\left( -1 \right)=3\left( -1 \right)-5\left( -1 \right)+11-3 \\
 & \Rightarrow p\left( -1 \right)=-3-5+11-3 \\
 & \Rightarrow p\left( -1 \right)=0 \\
\end{align}\]
Value of x as -1 gives p(x) as zero, therefore, -1 is a zero of the polynomial.
Putting $ x=\dfrac{-1}{3} $ in polynomial we get,
\[\begin{align}
  & p\left( \dfrac{-1}{3} \right)=3{{\left( \dfrac{-1}{3} \right)}^{3}}-5{{\left( \dfrac{-1}{3} \right)}^{2}}-11\left( \dfrac{-1}{3} \right)-3 \\
 & \Rightarrow p\left( \dfrac{-1}{3} \right)=3\left( \dfrac{-1}{27} \right)-5\left( \dfrac{1}{9} \right)+\dfrac{11}{3}-3 \\
 & \Rightarrow p\left( \dfrac{-1}{3} \right)=\dfrac{-1}{9}-\dfrac{5}{9}+\dfrac{11}{3}-3 \\
\end{align}\]
Taking LCM as 9 we get,
\[\begin{align}
  & \Rightarrow p\left( \dfrac{-1}{3} \right)=\dfrac{-1-5+33-27}{9} \\
 & \Rightarrow p\left( \dfrac{-1}{3} \right)=\dfrac{-33+33}{9} \\
 & \Rightarrow p\left( \dfrac{-1}{3} \right)=0 \\
\end{align}\]
Value of x as $ \dfrac{-1}{3} $ gives p(x) as zero, therefore, $ \dfrac{-1}{0} $ is a zero of the polynomial.
Hence all three zeroes are verified.
Now let us verify the relationship between zeroes and the coefficient.
We know that, for a cubic polynomial $ a{{x}^{3}}+b{{x}^{2}}+cx+d $ with zeroes as $ \alpha ,\beta ,\gamma $ we have,
\[\begin{align}
  & \alpha +\beta +\gamma =\dfrac{-b}{a} \\
 & \alpha \beta \gamma =\dfrac{-d}{a} \\
 & \alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{c}{a} \\
\end{align}\]
So here, $ \alpha =3,\beta =-1\text{ and }\gamma =\dfrac{-1}{3} $ .
Also comparing $ a{{x}^{3}}+b{{x}^{2}}+cx+d $ with $ 3{{x}^{3}}-5{{x}^{2}}-11x-3 $ we get, a = 3, b = -5, c = -11 and d = -3.
 $ \alpha +\beta +\gamma =3+\left( -1 \right)+\left( \dfrac{-1}{3} \right)\Rightarrow 3-1-\dfrac{1}{3} $ .
Taking LCM as 3 we get,
\[\alpha +\beta +\gamma =\dfrac{9-3-1}{3}=\dfrac{5}{3}\].
Also $ \dfrac{-b}{a}=\dfrac{-\left( -5 \right)}{3}=\dfrac{5}{3} $ .
Therefore, $ \alpha +\beta +\gamma =\dfrac{-b}{a} $ .
Now,
 $ \alpha \beta +\beta \gamma +\gamma \alpha =3\left( -1 \right)+\left( -1 \right)\left( \dfrac{-1}{3} \right)+\left( \dfrac{-1}{3} \right)\left( 3 \right)\Rightarrow -3+\dfrac{1}{3}-1\Rightarrow -4+\dfrac{1}{3} $ .
Taking LCM as 3 we get,
\[\alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{-12+1}{3}=\dfrac{-11}{3}\].
Also $ \dfrac{c}{a}=\dfrac{-11}{3} $ .
Therefore, \[\alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{c}{a}\].
Now \[\alpha \beta \gamma =\left( 3 \right)\left( -1 \right)\left( \dfrac{-1}{3} \right)=1\].
Also \[\dfrac{-d}{a}=\dfrac{-\left( -3 \right)}{3}=1\].
Therefore, \[\alpha \beta \gamma =\dfrac{-d}{a}\].

Note:
 Students should take care of the signs while solving this sum. Zeroes of a polynomial are also called roots of the polynomial. Students should remember the relationship between the zeroes and the coefficient of the polynomials.