Chapter 2.7, Problem 24E

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 $f\left(x\right)=x^3-2x^2+x-2$ .

Given information:

Zeros of a function: $2,-i,i$ .

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: For any polynomial function $f\left(x\right)$ with real coefficients, if $\left(a+bi\right)$ is an imaginary zero of the function, then $\left(a-bi\right)$ will be the other zero of $f\left(x\right)$ because complex zeros come in pairs.
3. As per Irrational conjugates Theorem, if $\left(a+\sqrt{b}\right)$ is a zero of a polynomial function, then $\left(a-\sqrt{b}\right)$ will be other zero of the function.

Calculation:

For the given problem, there are 3 zeros. This means there will be 3 factors of the given function as:

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

Now expand the factors using the distributive property $a\left(m+n\right)=a\cdot m+a\cdot n$ :

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

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