Chapter 4, Problem 29CR

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 29CR

The required polynomial is $f\left(x\right)=x^3-5x^2+11x-15$ .

Explanation of Solution

Given information:

The zeroes $=3,1+2i$

Calculation:

Consider the roots of the polynomial be $x_1=3$ , $x_2=1+2i$

Since, one of the roots of the polynomial is a complex number that is, $1+2i$ ;

So, the polynomial must have another root which is the conjugate of $1+2i$ that is, $x_3=1-2i$ .

The number of roots of the required polynomial is 3.

This means the smallest degree of the polynomial will be 3.

We can write the polynomial as:

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

Substitute 1 for a , 3 for $x_1$ , $1+2i$ for $x_2$ and $1-2i$ for $x_3$ in the above polynomial.

  $\begin{array}{c}f\left(x\right)=\left(x-3\right)\left[x-\left(1+2i\right)\right]\left[x-\left(1-2i\right)\right]\\ =\left(x-3\right)\left(x-1-2i\right)\left(x-1+2i\right)\\ =\left(x-3\right)\left[{\left(x-1\right)}^2-{\left(2i\right)}^2\right]\\ =\left(x-3\right)\left[{\left(x-1\right)}^2+4\right]\end{array}$

Simplify the above expression further.

  $\begin{array}{c}\left(x-3\right)\left[{\left(x-1\right)}^2+4\right]=\left(x-3\right)\left(x^2-2x+1+4\right)\\ =\left(x-3\right)\left(x^2-2x+5\right)\end{array}$

Expand the above expression.

  $\begin{array}{c}\left(x-3\right)\left(x^2-2x+5\right)=x^3-3x^2-2x^2+6x+5x-15\\ =x^3-5x^2+11x-15\end{array}$

Hence, the required polynomial is $f\left(x\right)=x^3-5x^2+11x-15$ .