Chapter 4, Problem 27CR

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 27CR

The required polynomial is $f\left(x\right)=x^4-9x^3+11x^2+51x-30$ .

Explanation of Solution

Given information:

The zeroes $=-2,5,3+\sqrt{6}$

Calculation:

Consider the roots of the polynomial be $x_1=-2$ , $x_2=5$ , $x_3=3+\sqrt{6}$

Since, one of the roots of the polynomial is a radical that is, $3+\sqrt{6}$ ;

So, the polynomial must have another root which is the conjugate of $3+\sqrt{6}$ that is, $x_4=3-\sqrt{6}$ .

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 $x_1$ , 5 for $x_2$ , $3+\sqrt{6}$ for $x_3$ and $3-\sqrt{6}$ for $x_4$ in the above polynomial.

  $\begin{array}{c}f\left(x\right)=1\left(x+2\right)\left(x-5\right)\left[x-\left(3+\sqrt{6}\right)\right]\left[x-\left(3-\sqrt{6}\right)\right]\\ =\left(x^2+2x-5x-10\right)\left[x-3-\sqrt{6}\right]\left[x-3+\sqrt{6}\right]\\ =\left(x^2-3x-10\right)\left[{\left(x-3\right)}^2-{\left(\sqrt{6}\right)}^2\right]\\ =\left(x^2-3x-10\right)\left[x^2-6x+9-6\right]\end{array}$

Simplify the above expression further.

  $\left(x^2-3x-10\right)\left[x^2-6x+9-6\right]=\left(x^2-3x-10\right)\left(x^2-6x+3\right)$

Expand the above expression.

  $\begin{array}{c}\left(x^2-3x-10\right)\left(x^2-6x+3\right)=x^4-6x^3+3x^2-3x^3+18x^2-9x-10x^2+60x-30\\ =x^4-9x^3+11x^2+51x-30\end{array}$

Hence, the required polynomial is $f\left(x\right)=x^4-9x^3+11x^2+51x-30$ .