Question

How do you write a polynomial function with the given zeros 2 ,$2i$ and degree 3?

Answer 1

Hint: We use the factor theorem to conclude that if $x=\alpha ,\beta ,\gamma $ are the roots of a polynomial of third degree (cubic) then the polynomial is given by $\left( x-\alpha \right)\left( x-\beta \right)\left( x-\gamma \right)$. We recall the nature of zeroes of cubic polynomials that if one zero is real then other two will be complex conjugates. We use it to get the third zero as $x=-2i$.

Complete step by step answer:
We know for any polynomial $p\left( x \right)$ if for some $x=a$ we have $p\left( a \right)=0$ then we call $x=a$ a zero or root of the polynomial$p\left( x \right)$. If the degree of polynomial is $n$ then we have $n$zeroes. Since we have to make a polynomial of degree 3 it must have three zeros. Let us denote the zeros of the required degree 3 (also known as cubic) polynomial as $x=\alpha ,\beta ,\gamma $. \[\]
We know from factor theorem that if $x=a$ is zero of $p\left( x \right)$ then $x-a$ is a factor of $p\left( x \right)$. Since we have three zeros $x=\alpha ,\beta ,\gamma $ we shall have exactly three factors $x-\alpha ,x-\beta ,x-\gamma $ and can express the polynomial as
\[p\left( x \right)=\left( x-\alpha \right)\left( x-\beta \right)\left( x-\gamma \right)\]
We are given in the question two zeros which we denote as $\alpha =2,\beta =2i$. We know from the nature of zeros cubic polynomials that we can either get three real zeros or one real zero and two conjugate complex zeros. We know that complex zeros are in the form $a+bi,a-bi$. Here we have one complex zero $a+bi=2i\Rightarrow 0+bi=0+2i\Rightarrow a=0,b=2$. Then the other zero is
\[\gamma =a-bi=0-2i=-2i\]
So the third factor is
\[\left( x-\left( -2i \right) \right)=\left( x+2i \right)\]
So the required degree 3 polynomial is
\[\begin{align}
  & p\left( x \right)=\left( x-2 \right)\left( x-2i \right)\left( x+2i \right) \\
 & \Rightarrow p\left( x \right)=\left( x-2 \right)\left( {{x}^{2}}-{{\left( 2i \right)}^{2}} \right) \\
 & \Rightarrow p\left( x \right)=\left( x-2 \right)\left( {{x}^{2}}-\left( -4 \right) \right) \\
 & \Rightarrow p\left( x \right)=\left( x-2 \right)\left( {{x}^{2}}+4 \right) \\
 & \Rightarrow p\left( x \right)={{x}^{3}}-2{{x}^{2}}-4x-8 \\
\end{align}\]

Note:
We have used the identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ and the definition ${{i}^{2}}=-1$above. We note that the general cubic equation is $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$ whose discriminant is given by $D=abcd-4{{b}^{3}}d+{{b}^{2}}{{c}^{2}}-4a{{c}^{3}}-27{{a}^{2}}{{d}^{2}}$. If $D\ge 0$ we get three real roots and if $D=0$ we get at least one repeated root. If $D<0$ we get one real and two conjugate complex roots.