Question

Given that $x-\sqrt{5}$ is a factor of cubic polynomial ${{x}^{3}}+3\sqrt{5}{{x}^{2}}+13x-3\sqrt{5}$, find all the zeroes of the polynomial.

Answer 1

Hint: We will divide the given polynomial by the given factor, i.e. $x-\sqrt{5}$ . The quotient obtained will be a quadratic. We will then factorize the quadratic. Then, we will use the fact that if (x – a) is a factor of the polynomial p(x), then “a” will be a zero of the polynomial p(x).

Complete step-by-step answer:
The given polynomial in the question is ${{x}^{3}}+3\sqrt{5}{{x}^{2}}+13x-3\sqrt{5}$. Here, highest power of x is 3, that is the order of polynomial is 3. So, it will have three zeroes, out of which one zero we can get from given factor by equating it with o.
So, one of the zero will be given as,
$\begin{align}
  & x-\sqrt{5}=0 \\
 & \Rightarrow x=\sqrt{5} \\
\end{align}$
Let the remaining two zero of given polynomial be $\alpha $ and $\beta $. So, corresponding factors of zeroes $\alpha $ and $\beta $ will be $\left( x-\alpha \right)$and $\left( x-\beta \right)$. We can write the given polynomial as product of these three factors.
Taking k to be a constant number, we can write,
$k\left( x-\sqrt{5} \right)\left( x-\alpha \right)\left( x-\beta \right)={{x}^{3}}-3\sqrt{5}{{x}^{2}}+13x-3\sqrt{5}$
Dividing both sides of equation with, we get,
\[k\left( x-\alpha \right)\left( x-\beta \right)=\dfrac{{{x}^{3}}-3\sqrt{5}{{x}^{2}}+13x-3\sqrt{5}}{x-\sqrt{5}}.........(i)\]
Let us divide right hand side of this equation using long division method,
So, we get,
\[\begin{align}
  & x-\sqrt{5}\overset{{{x}^{2}}-2\sqrt{5}x+3}{\overline{\left){{{x}^{3}}-3\sqrt{5}{{x}^{2}}+13x-3\sqrt{5}}\right.}} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{3}}-\sqrt{5}{{x}^{2}} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline{\begin{align}
  & \,\,\,\,\,\,\,\,\,-2\sqrt{5}{{x}^{2}}+13x-3\sqrt{5} \\
 & \,\,\,\,\,\,\,\,-2\sqrt{5}{{x}^{2}}\,+13x-\,\,3\sqrt{5} \\
 & \overline{\begin{align}
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3x-3\sqrt{5} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3x-3\sqrt{5} \\
 & \,\overline{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,} \\
\end{align}} \\
\end{align}} \\
\end{align}\]
So, on dividing, we get the quotient to be \[{{x}^{2}}-2\sqrt{5}x+3\] and remainder 0.
Using this value of the quotient in equation (i) we get,
\[k\left( x-\alpha \right)\left( x-\beta \right)={{x}^{2}}-2\sqrt{5}x+3\]
Right hand side of this equation is the quadratic equation. Let us use quadratic formula \[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] to solve this, so, we get,
\[\begin{align}
  & x=\dfrac{-\left( -2\sqrt{5} \right)\pm \sqrt{{{\left( -2\sqrt{5} \right)}^{2}}-4\times 1\times 3}}{2\times 1} \\
 & \Rightarrow \dfrac{2\sqrt{5}\sqrt{4\times 5-12}}{2} \\
 & \Rightarrow \dfrac{2\sqrt{5}\pm \sqrt{8}}{2} \\
\end{align}\]
Here, we can write \[\sqrt{8}\] as \[2\sqrt{2}\],
\[x=\dfrac{2\sqrt{5}\pm 2\sqrt{2}}{2}\]
Therefore, when \[x=\dfrac{2\sqrt{5}+2\sqrt{2}}{2}\] or \[x=\dfrac{2\sqrt{5}-2\sqrt{2}}{2}\]
\[\Rightarrow x=\dfrac{2\left( \sqrt{5}+\sqrt{2} \right)}{2}\] or \[x=\dfrac{2\left( \sqrt{5}-\sqrt{2} \right)}{2}\]
Therefore, \[\alpha =\sqrt{5}+\sqrt{2}\] and \[\beta =\sqrt{5}-\sqrt{2}\].
Hence, all the zeros of a given polynomial are \[\sqrt{5\,},\,\sqrt{5+2}\,\] and \[\sqrt{5-2}\].

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.