Question

How do you find a fourth degree polynomial given roots $3i$ and $\sqrt{6}$?

Answer 1

Hint: The roots of the fourth degree polynomial are $3i$ and $\sqrt{6}$, which are respectively the imaginary and the irrational roots. The other two roots of the fourth degree polynomial can be found by using the fact that the irrational and the imaginary roots always occur in the conjugate pairs. So the other roots of the polynomial will be $-3i$ and $-\sqrt{6}$. Using the factor theorem, we will be able to generate all the four factors of the polynomial, and finally on multiplying these, we will obtain the fourth degree polynomial.

Complete step by step solution:
According to the above question, the polynomial is having a degree of four whose roots are $3i$ and $\sqrt{6}$. The fourth degree of the polynomial means that the roots must also be four. We can see that the first given root is imaginary, while the second given root is irrational. We know that the imaginary and the irrational roots always occur in the conjugate pairs. This means that the other two roots of the polynomial must be $-3i$ and $-\sqrt{6}$.
So the roots of the fourth degree polynomial are $3i$, $-3i$, $\sqrt{6}$ and $-\sqrt{6}$. By the factor theorem we know that if $a$ is a root of a polynomial $p\left( x \right)$, then $\left( x-a \right)$ is a factor of $p\left( x \right)$. Therefore, from the roots $3i$, $-3i$, $\sqrt{6}$ and $-\sqrt{6}$ of the fourth degree polynomial, we get the factors as $\left( x-3i \right)$, $\left( x+3i \right)$, \[\left( x-\sqrt{6} \right)\] and \[\left( x+\sqrt{6} \right)\] respectively. So the fourth degree polynomial can be given by
$\Rightarrow p\left( x \right)=\left( x-3i \right)\left( x+3i \right)\left( x-\sqrt{6} \right)\left( x+\sqrt{6} \right)$
Using the algebraic identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ we get
\[\begin{align}
  & \Rightarrow p\left( x \right)=\left( {{x}^{2}}-{{\left( 3i \right)}^{2}} \right)\left( {{x}^{2}}-{{\sqrt{6}}^{2}} \right) \\
 & \Rightarrow p\left( x \right)=\left( {{x}^{2}}-9{{i}^{2}} \right)\left( {{x}^{2}}-6 \right) \\
\end{align}\]
Now, we know that \[{{i}^{2}}=-1\]. Putting this in the above polynomial, we get
\[\begin{align}
  & \Rightarrow p\left( x \right)=\left( {{x}^{2}}-9\left( -1 \right) \right)\left( {{x}^{2}}-6 \right) \\
 & \Rightarrow p\left( x \right)=\left( {{x}^{2}}+9 \right)\left( {{x}^{2}}-6 \right) \\
\end{align}\]
Now, using the distributive law of the algebraic multiplication, we get
\[\begin{align}
  & \Rightarrow p\left( x \right)=\left( {{x}^{2}} \right)\left( {{x}^{2}}-6 \right)+9\left( {{x}^{2}}-6 \right) \\
 & \Rightarrow p\left( x \right)={{x}^{4}}-6{{x}^{2}}+9{{x}^{2}}-54 \\
 & \Rightarrow p\left( x \right)={{x}^{4}}+3{{x}^{2}}-54 \\
\end{align}\]
Hence, the fourth degree polynomial is found as \[{{x}^{4}}+3{{x}^{2}}-54\].

Note:
In the above question, only two roots out of the four possible roots of the fourth degree polynomial were given so as to judge our concept of conjugate roots. The concept of the conjugate roots is evident from the $\pm $ sign in the quadratic formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. It is well known that the complex roots occur in conjugate pairs, but students forget that the irrational roots also occur in conjugate pairs. If $A+\sqrt{B}$ is a root of a polynomial, then $A-\sqrt{B}$ will also be a root of the polynomial.