Chapter 2, Problem 20T

Solution

To write: A polynomial function $f$ of least degree that has rational coefficients, a leading coefficient of $1$ and the given zeros.

The required polynomial function would be $f\left(x\right)=x^3-6x^2+4x-24$ .

Given information:

Zeros of a function: $6,\text{}2i$ .

Formula used:

1. Factor Theorem: If $x=a,b,c$ are zeros of the function $f\left(x\right)$ , then $\left(x-a\right),\left(x-b\right)\text{and }\left(x-c\right)$ would be factors of the function.
2. Complex Conjugate Theorem: If $f$ is a polynomial function with real coefficients, and $\left(a+bi\right)$ is an imaginary zero of $f$ , then $\left(a-bi\right)$ is also a zero of $f$ .
3. Irrational conjugates Theorem: Suppose $f$ is a polynomial function with real coefficients, and $a\text{and }b$ are rational numbers such that $\sqrt{b}$ is irrational. If $a+\sqrt{b}$ is a zero of $f$ , then $a+\sqrt{b}$ is also a zero of $f$ .

Calculation:

For the given problem, there are $2$ zeros. As per complex conjugate theorem, the other zero would be $-2i$ .

Using the factor theorem, the factored form of $f$

would be:

  $\begin{array}{c}f\left(x\right)=\left(x-\left(-2i\right)\right)\left(x-2i\right)\left(x-6\right)\\ =\left(x+2i\right)\left(x-2i\right)\left(x-6\right)\end{array}$

Now expand the factors using the difference of squares $\left(a-b\right)\left(a+b\right)=a^2-b^2$:

  $\begin{array}{c}f\left(x\right)=\left(x+2i\right)\left(x-2i\right)\left(x-6\right)\\ =\left(x^2-{\left(2i\right)}^2\right)\left(x-6\right)\\ =\left(x^2-4i^2\right)\left(x-6\right)\\ =\left(x^2+4\right)\left(x-6\right)\left(\text{Using identity }-i^2=1\right)\\ =x^2\left(x-6\right)+4\left(x-6\right)\\ =x^3-6x^2+4x-24\end{array}$

Therefore, the required function would be $f\left(x\right)=x^3-6x^2+4x-24$ .