Question

How do you find an example of a fourth degree of polynomial equation that has no real zeros?

Answer 1

Hint: In the above question we are asked to find a fourth-degree polynomial which has no real roots. This means the roots are imaginary. In order to find the polynomial, pick any conjugate pair of imaginary numbers. Then by simply multiplying them we will get the required fourth degree of polynomial equation.

Complete step by step answer:
Now here we have to find biquadratic polynomials by assuming two imaginary roots with their conjugates as it has no real roots therefore, the roots of the biquadratic polynomial will be imaginary.
We will randomly take any complex numbers with their conjugates as imaginary roots always occur in pairs.
Therefore, to find an example of a fourth degree polynomial equation that has no real zeros or only has imaginary roots, we will assume \[1\pm i\] and \[2\pm i\] as the roots.
So, the polynomial will be,
\[\begin{align}
  & f(x)=(x-(1+i))(x-(1-i))(x-(2+i))(x-(1-i)) \\
 & \Rightarrow f(x)=((x-1)-i)((x-1)+i)((x-1)-i)((x-2)+i) \\
 & \Rightarrow f(x)=({{(x-1)}^{2}}-{{i}^{2}})({{(x-2)}^{2}}-{{i}^{2}}) \\
 & \Rightarrow f(x)=({{(x-1)}^{2}}+1)({{(x-2)}^{2}}+1) \\
 & \Rightarrow f(x)=({{x}^{2}}-2x+2)({{x}^{2}}-4x+5) \\
 & \therefore f(x)=({{x}^{4}}-6{{x}^{3}}+15{{x}^{2}}-18x+10) \\
\end{align}\]
After simplifying we get \[f(x)=({{x}^{4}}-6{{x}^{3}}+15{{x}^{2}}-18x+10)\] .
Therefore, an example of a fourth degree of polynomial equation that has no real zeros is \[{{x}^{4}}-6{{x}^{3}}+15{{x}^{2}}-18x+10\] .

Note: The other method to solve the question is that we can start with any quadratic equation which has a positive leading coefficient or the discriminant value of that quadratic equation is less than zero and then constantly increasing the constant term till the value where it no longer intersects the x-axis. While solving the above equation keep in mind to take the imaginary roots in conjugate pairs.