Chapter 4, Problem 26CR

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 26CR

The required polynomial is $f\left(x\right)=x^4-5x^3+x^2+25x-30$ .

Explanation of Solution

Given information:

The zeroes $=2,3,\sqrt{5}$

Calculation:

Consider the roots of the polynomial be $x_1=2$ , $x_2=3$ , $x_3=\sqrt{5}$

Since, one of the roots of the polynomial is a radical that is, $\sqrt{5}$ ;

So, the polynomial must have another root which is the conjugate of $\sqrt{5}$ that is,

  $x_4=-\sqrt{5}$ .

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 , $2$ for $x_1$ , 3 for $x_2$ , $\sqrt{5}$ for $x_3$ and $-\sqrt{5}$ for $x_4$ in the above polynomial.

  $\begin{array}{c}f\left(x\right)=1\left(x-2\right)\left(x-3\right)\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\\ =\left(x^2-3x-2x+6\right)\left[x^2-{\left(\sqrt{5}\right)}^2\right]\\ =\left(x^2-5x+6\right)\left(x^2-5\right)\end{array}$

Expand the above expression.

  $\begin{array}{c}\left(x^2-5x+6\right)\left(x^2-5\right)=x^4-5x^2-5x^3+25x+6x^2-30\\ =x^4-5x^3+x^2+25x-30\end{array}$

Hence, the required polynomial is $f\left(x\right)=x^4-5x^3+x^2+25x-30$ .