Chapter 4, Problem 30CR

To determine

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

Answer to Problem 30CR

The required polynomial is $f\left(x\right)=x^4-x^3+14x^2-16x-32$ .

Explanation of Solution

Given information:

The zeroes $=-1,2,4i$

Calculation:

Consider the roots of the polynomial be $x_1=-1$ , $x_2=2$ , $x_3=4i$

Since, one of the roots of the polynomial is a complex number that is, $4i$ ;

So, the polynomial must have another root which is the conjugate of $4i$ that is, $x_4=-4i$ .

The number of roots of the required polynomial is 4.

This means the smallest degree of the polynomial will be 4.

We can write the polynomial as:

  $f\left(x\right)=a\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right)$ , where $a$ is the leading coefficient.

Substitute 1 for a , $-1$ for $x_1$ , $2$ for $x_2$ , $4i$ for $x_3$ and $-4i$ for $x_4$ in the above polynomial.

  $\begin{array}{c}f\left(x\right)=\left(x+1\right)\left(x-2\right)\left(x-4i\right)\left(x+4i\right)\\ =\left(x^2+x-2x-2\right)\left[x^2-{\left(4i\right)}^2\right]\\ =\left(x^2-x-2\right)\left(x^2+16\right)\end{array}$

Expand the above expression.

  $\begin{array}{c}\left(x^2-x-2\right)\left(x^2+16\right)=x^4-x^3-2x^2+16x^2-16x-32\\ =x^4-x^3+14x^2-16x-32\end{array}$

Hence, the required polynomial is $f\left(x\right)=x^4-x^3+14x^2-16x-32$ .