Question

Find all zeros of the polynomial $2{{x}^{3}}+{{x}^{2}}-6x-3$, if two of its zeroes are $\sqrt{3}$ and \[-\sqrt{3}\].

Answer 1

Hint:We will use the fact that if “a” is a zero of the polynomial p(x), then (x – a) will be a factor of the polynomial p(x). So, we will divide the given polynomial by $\left( x-\sqrt{3} \right)\left( x+\sqrt{3} \right)$ . The quotient will be a linear equation in x. We will solve the linear equation to find the other zero

Complete step-by-step answer:
In a given question, we have a polynomial $2{{x}^{3}}+{{x}^{2}}-6x-3$, where highest power of $x$ is $3$, that is the polynomial is of order $3$. So, it will have three zeroes.
Now, two of the zeroes of the given polynomial are given in question, which are $\sqrt{3}$ and \[-\sqrt{3}\] .
Let us consider third zero of these polynomials to be $\alpha $. Then, we can write this polynomial as a product of factors of its zero, that is $\left( x-\sqrt{3} \right),\,\left( x-\left( -\sqrt{3} \right) \right)$ and $\left( x-\alpha \right)$. Therefore, taking $k$ to be any constant number, we can write,
\[\begin{align}
  & k\left( x-\sqrt{3} \right)\left( x-\left( -\sqrt{3} \right) \right)\left( x-\alpha \right)=2{{x}^{3}}+{{x}^{2}}-6x-3 \\
 & \Rightarrow k\left( x-\sqrt{3} \right)\left( x+\sqrt{3} \right)\left( x-\alpha \right)=2{{x}^{3}}+{{x}^{2}}-6x-3 \\
\end{align}\]
Using formula, $\left( a+b \right)\left( a-b \right)={{a}^{2}}+{{b}^{2}}$, we can write the above equation as,
$\begin{align}
  & k\left( {{x}^{2}}-{{\left( \sqrt{3} \right)}^{2}} \right)\left( x-\alpha \right)=2{{x}^{2}}+{{x}^{2}}-6x-3 \\
 & \Rightarrow k\left( {{x}^{2}}-3 \right)\left( x-\alpha \right)=2{{x}^{2}}+{{x}^{2}}-6x-3 \\
\end{align}$
Dividing both sides of this equation with ${{x}^{2}}-3$, we get,
\[k\left( x-\alpha \right)=\dfrac{2{{x}^{3}}+{{x}^{2}}-6x-3}{{{x}^{2}}-3}..............(i)\]
Let us divide right hand side of this equation using long division method as below,
\[\]

\[{{x}^{2}}-3\overset{2x-1}{\overline{\left){\begin{align}
  & 2{{x}^{3}}+{{x}^{2}}-6x-3 \\
 & 2{{x}^{3}}\,\,\,\,\,\,\,\,\,\,-6x \\
 & \overline{\begin{align}
  & \,\,\,\,\,\,\,\,\,{{x}^{2}}+0-3 \\
 & \,\,\,\,\,\,\,\,\,{{x}^{2}}\,\,\,\,\,\,\,\,-3 \\
 & \overline{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,}\,\, \\
\end{align}} \\
\end{align}}\right.}}\]
So, on dividing, we get the quotient to be $2x+1$ and the remainder $0$. Using this value of quotient in equation(i), we get,
$k\left( x-2 \right)=2x+1$
Taking 2 common from right hand side of this equation, we get,
$k\left( x-\alpha \right)=2\left( x+\dfrac{1}{2} \right)$
Comparing both sides of this equation, \[k=2,\,\,\alpha =-\dfrac{1}{2}=-0.5\].
Hence, all the zeros of given polynomial are $\sqrt{3\,},\,-\sqrt{3}$ and $-0.5$

Note: While doing calculations, be careful of the sign. Sign mistakes are very common and hence, students should be very careful while doing calculations and substitutions.Students should also know about the calculation procedure of long division method for solving these types of questions.