Question

Given that $\sqrt{2}$ is a zero of the cubic equation $6{{x}^{3}}+\sqrt{2}{{x}^{2}}-10x-4\sqrt{2}$, find its other two zero.

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 $\left( x-\sqrt{2} \right)$ . The quotient will be a quadratic in x. We will solve the quadratic to find the other zeroes.

Complete step-by-step answer:
In the given question, we have a polynomial, $6{{x}^{3}}+\sqrt{2}{{x}^{2}}-10x-4\sqrt{2}$, where highest power of x is 3, that is order of this polynomial is 3. So, here we will have 3 zeroes, out of which one zero is already given in a question, which is $\sqrt{2}$ .
Let us consider the remaining two zeroes of this polynomial to be $\alpha $ and $\beta $.
Hence, this polynomial will have 3 factors from which we will get this three zero.
So, we can write this polynomial as a product of three factors. Taking k to be any constant number, we can write,
\[\] $k\left( x-\sqrt{2} \right)\left( x-\alpha \right)\left( x-\beta \right)=6{{x}^{3}}+\sqrt{2}{{x}^{2}}-10x-4\sqrt{2}$
Dividing on both sides of this equation with $x-\sqrt{2}$, we get,
\[k\left( x-\sqrt{2} \right)\left( x-\alpha \right)\left( x-\beta \right)=\dfrac{6{{x}^{3}}+\sqrt{2}{{x}^{2}}-10x-4\sqrt{2}}{x-\sqrt{2}}.........(i)\]
Let us divide right hand side of this equation using long division method, as given below,
\[\begin{align}
  & x-\sqrt{2}\overset{6{{x}^{2}}+7\sqrt{2}x+4}{\overline{\left){6{{x}^{3}}+\sqrt{2}{{x}^{2}}-10x-4\sqrt{2}}\right.}}\,\,\,\, \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,6{{x}^{3}}\,\,\,\,6\sqrt{2}{{x}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline{\begin{align}
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7\sqrt{2}{{x}^{2}}-10x-4\sqrt{2} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7\sqrt{2}{{x}^{2}}-14x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
 & \overline{\begin{align}
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4x-4\sqrt{2} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4x-4\sqrt{2} \\
 & \overline{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
\end{align}}\, \\
\end{align}} \\
\end{align}\]
 So, on dividing we get, quotient to be \[6{{x}^{2}}+7\sqrt{2}x+4\] and remainder 0.
Using this value of quotient in equation (i), we get, $k\left( x-\alpha \right)\left( x-\beta \right)=6{{x}^{2}}+7\sqrt{2}x+4$.
Splitting the middle term of right-hand side expression, to factorise it, we get,
$k\left( x-\alpha \right)\left( x-\beta \right)=6{{x}^{2}}+4\sqrt{2}x+3\sqrt{2}x+4$.
Here, we can write $6{{x}^{2}}$as, $6{{x}^{2}}=3\times 2{{x}^{2}}$, here 2 can be written as product of $\sqrt{2}$ and $\sqrt{2}$. So, we can write, $6{{x}^{2}}-3\sqrt{2}\times \sqrt{2}x$
Using this in above equation, we can write,
 $k\left( x-\alpha \right)\left( x-\beta \right)=\sqrt{2}\times 3\sqrt{2}{{x}^{2}}+4\sqrt{2}x+3\sqrt{2}x+4$
Taking common terms out, we can write,
$k\left( x-\alpha \right)\left( x-\beta \right)=\sqrt{2}x\left( 3\sqrt{2}x+4 \right)+1\left( 3\sqrt{2}x+4 \right)$
Taking $\left( 3\sqrt{2}x+4 \right)$ common, we get,
$k\left( x-\alpha \right)\left( x-\beta \right)=\left( 3\sqrt{2}x+4 \right)\left( \sqrt{2}x+4 \right)$
Taking \[3\sqrt{2}\] common from first factor and $\sqrt{2}$ common from second factor of right hand side, we get,
\[\] $\begin{align}
  & k\left( x-\alpha \right)\left( x-\beta \right)=3\sqrt{2}\left( x+\dfrac{4}{3\sqrt{2}} \right)\sqrt{2}\left( x+\dfrac{1}{\sqrt{2}} \right) \\
 & \Rightarrow 3\sqrt{2}\times \sqrt{2}\left( x+\dfrac{4}{3\sqrt{2}} \right)\left( x+\dfrac{1}{\sqrt{2}} \right) \\
 & \Rightarrow 3\times 2\left( x+\dfrac{4}{3\sqrt{2}} \right)\left( x+\dfrac{1}{\sqrt{2}} \right) \\
\end{align}$
Simplifying, this we get,
 \[\begin{align}
  & k\left( x-\alpha \right)\left( x-\beta \right)=6\left( x+\dfrac{2\sqrt{2}}{3} \right)\sqrt{2}\left( x+\dfrac{\sqrt{2}}{2} \right) \\
 & \Rightarrow 6\left( x-\left( \dfrac{-2\sqrt{2}}{3} \right) \right)\sqrt{2}\left( x-\left( \dfrac{-\sqrt{2}}{2} \right) \right) \\
\end{align}\]
Compounding both sides of equation, we get, $k=6,\,\,\alpha =\dfrac{-2\sqrt{2}}{3}$and $\beta =\dfrac{-\sqrt{2}}{2}$.
Hence, the other two zero of given polynomial are \[\dfrac{-2\sqrt{2}}{2}\] and \[\dfrac{-\sqrt{2}}{2}\].

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.