Chapter 13, Problem 5CR

To determine

To find: The complex zeros of $f\left(x\right)=5x^4-9x^3-7x^2-31x-6$ .

Answer to Problem 5CR

  $\frac{-1+\sqrt{7}i}{2}\text{ and }\frac{-1-\sqrt{7}i}{2}$

Explanation of Solution

Given: $f\left(x\right)=5x^4-9x^3-7x^2-31x-6$

The given function is $f\left(x\right)=5x^4-9x^3-7x^2-31x-6$ .

This is fourth degree polynomial. First root find using hit-trial error method.

Put $x=3$ into $f\left(x\right)$

  $f\left(3\right)=5{\left(3\right)}^4-9{\left(3\right)}^3-7{\left(3\right)}^2-31\left(3\right)-6=0$

Thus, $x=3$ is a zero of the polynomial.

Put $x=-\frac{1}{5}$ into $f\left(x\right)$

  $f\left(-\frac{1}{5}\right)=5{\left(-\frac{1}{5}\right)}^4-9{\left(-\frac{1}{5}\right)}^3-7{\left(-\frac{1}{5}\right)}^2-31\left(-\frac{1}{5}\right)-6=0$

Thus, $x=-\frac{1}{5}$ is a zero of the polynomial.

So, factor of $f\left(x\right)$ is $\left(x-3\right)\left(x+\frac{1}{5}\right)=x^2-\frac{14}{5}x-\frac{3}{5}$

Now divide the polynomial by $x^2-\frac{14}{5}x-\frac{3}{5}$

  $\frac{5x^4-9x^3-7x^2-31x-6}{x^2-\frac{14}{5}x-\frac{3}{5}}=5x^2+5x+10$

Now factor $5x^2+5x+10$ using quadratic formula.

  $\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2ac}\left[\text{Quadratic formula}\right]\\ x=\frac{-5\pm \sqrt{5^2-4\left(5\right)\left(10\right)}}{2\left(5\right)}\\ x=\frac{-5\pm \sqrt{-175}}{10}\\ x=\frac{-1\pm \sqrt{7}i}{2}\end{array}$

Hence, the complex zeros are $\frac{-1+\sqrt{7}i}{2}\text{ and }\frac{-1-\sqrt{7}i}{2}.$