Chapter 6.7, Problem 57HP

To determine

To write: The equation I factored form of a polynomial function of degree 5 with 2 imaginary zeros, 1 non-integer zero, and 2 irrational zeros.

Answer to Problem 57HP

The equation is $f\left(x\right)=4\left(x-i\right)\left(x+i\right)\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\left(x-\frac{5}{2}\right)$ .

Explanation of Solution

Given:

It is give that the function has 5 with 2 imaginary zeros, 1 non-integer zero, and 2 irrational zeros.

Calculation:

As the function has 5 degree so, the function will in the following form.

  $f\left(x\right)=a\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right)\left(x-x_5\right)$

Where, $x_1,x_2$ are imaginary, $x_3$ is non-integer and $x_4,x_5$ are irrational.

It is not given that the function should have rational coefficients, but for simplicity, look for an example with rational coefficients This means that the imaginary zeros will come in pairs of conjugates and the same will go for the irrational zeros

For example:.

  $\begin{array}{c}{x}_1=i\\ x_2=-i\\ x_4=\sqrt{3}\\ x_5=-\sqrt{3}\end{array}$

Assume $x_3$ as a non-integer value and assume an arbitrary constant $a$ as follows.

  $\begin{array}{c}{x}_3=\frac{5}{3}\\ a=4\end{array}$

Substitute all these values in the standard equation.

  $f\left(x\right)=4\left(x-i\right)\left(x+i\right)\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\left(x-\frac{5}{2}\right)$

Therefore, the equation is $f\left(x\right)=4\left(x-i\right)\left(x+i\right)\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\left(x-\frac{5}{2}\right)$ .