Question

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

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 $\left( x-\sqrt{2} \right)\left( x+\sqrt{2} \right)$ . 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 this question, the given polynomial is ${{x}^{3}}+3{{x}^{2}}-2x-6$, which has highest power of x three, that is order of this polynomial is three, out of which two zeroes are already given in question. let us consider the third zero to be $\alpha $. Then this polynomial can be written as a product of three factors, each containing one of the zeroes of the polynomial. Here, the factors are $\left( x-\sqrt{2} \right),\,\left( x-\left( -\sqrt{2} \right) \right),\,\left( x-\alpha \right)$.
Taking k to be any constant number, we can write the given polynomial as,
$\begin{align}
  & k\left( x-\sqrt{2} \right)\left( x-\left( -\sqrt{2} \right) \right)\left( x-\alpha \right)={{x}^{3}}+3{{x}^{2}}-2x-6 \\
 & \Rightarrow k\left( x-\sqrt{2} \right)\left( x+\sqrt{2} \right)\left( x-\alpha \right)={{x}^{3}}+3{{x}^{2}}-2x-6 \\
\end{align}$
Using formula $\left( a=b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ here, we get,
$\begin{align}
  & k\left( {{x}^{2}}-{{\left( \sqrt{2} \right)}^{2}} \right)\left( x-\alpha \right)={{x}^{3}}+3{{x}^{2}}-2x-6 \\
 & \Rightarrow k\left( {{x}^{2}}-2 \right)\left( x-\alpha \right)={{x}^{3}}+3{{x}^{2}}-2x-6 \\
\end{align}$
Dividing both sides of this equation with $\left( {{x}^{2}}-2 \right)$, we get,
$k\left( x-\alpha \right)=\dfrac{{{x}^{3}}+3{{x}^{2}}-2x+6}{{{x}^{2}}-2}........(i)$
Let us divide right hand side of this equation using long division method, as below,
$\begin{align}
  & {{x}^{2}}-2\overset{x+3}{\overline{\left){{{x}^{3}}+3{{x}^{2}}-2x-6}\right.}} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-2x \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\overline{\begin{align}
  & \,\,\,\,\,\,\,\,\,\,\,3{{x}^{2}}+0\,\,\,\,-6 \\
 & \,\,\,\,\,\,\,\,\,\,\,3{{x}^{2}}\,\,\,\,\,\,\,\,\,\,\,-6 \\
 & \overline{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,}\,\, \\
\end{align}} \\
\end{align}$
 So, on dividing, we get, quotient to be x+3 and remainder 0.
Using this value of quotient in equation (i), we get,
$\begin{align}
  & k\left( x-\alpha \right)=x+3 \\
 & \Rightarrow k\left( x-\alpha \right)=1\left( x-\left( -3 \right) \right) \\
\end{align}$
Compounding both sides of this equation we get, $k=1,\,\alpha =-3$
Hence all zeros of a given polynomial are $\sqrt{2,\,}\,-\sqrt{2},\,-3$.

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.