Chapter 4, Problem 25CR

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 25CR

The required polynomial is $f\left(x\right)=x^3-5x^2+5x-1$ .

Explanation of Solution

Given information:

The zeroes $=1,2-\sqrt{3}$

Calculation:

Consider the roots of the polynomial be $x_1=1$ , $x_2=2-\sqrt{3}$ .

Since, one of the roots of the polynomial is a radical that is, $2-\sqrt{3}$ ;

So, the polynomial must have another root which is the conjugate of $2-\sqrt{3}$ that is, $x_3=2+\sqrt{3}$ .

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 , $1$ for $x_1$ , $2-\sqrt{3}$ for $x_2$ and $2+\sqrt{3}$ for $x_3$ in the above polynomial.

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

Simplify the above expression further.

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

Expand the above expression.

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

Hence, the required polynomial is $f\left(x\right)=x^3-5x^2+5x-1$ .