Chapter 13, Problem 5CR

To determine

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

Answer to Problem 5CR

Solution:

The complex zeros of $f\left(x\right)=5x^4-9x^3-7x^2-31x-6$ are $\left\{-\frac{1}{2}+\frac{\sqrt{7}}{2}i,-\frac{1}{2}-\frac{\sqrt{7}}{2}i,-\frac{1}{5},3\right\}$.

Explanation of Solution

Given information:

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

Formula used:

The solution of the quadratic equation $a{x}^2+bx+c=0$ is $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Calculation:

The degree of $f\left(x\right)$ is $4$. Therefore, $f\left(x\right)$ has four complex zeros.

By using the Descartes Rule of signs to find the real zeros,

$f\left(x\right)=5x^4-9x^3-7x^2-31x-6$, and $f\left(-x\right)=5x^4+9x^3-7x^2+31x-6$.

The number of positive real zeros of $f\left(x\right)$ are $1$, as changes in the sign of the coefficient of $f\left(x\right)$ are $1$.

The number of negative real zeros of $f\left(x\right)$ are $3$, since changes in the sign of the coefficient of $f\left(-x\right)$ are $3$.

By using the Rational zeros Theorem to find the potential rational zeros of $f\left(x\right)$,

The factor of $6$; all possible values of $p$ are $\pm 1,\pm 2,\pm 3,\pm 6$.

The factor of $5$; all possible value of $q$ are $\pm 1,\pm 5$.

The potential rational zeros are $\pm 1,\pm \frac{1}{5},\pm 2,\pm \frac{2}{5},\pm 3,\pm \frac{3}{5},\pm 6,\pm \frac{6}{5}$.

$\begin{array}{ccc}r& coefficientofquotient& remainder\\ 1& 5-4-11-42& -48\\ -1& 5-9-2-40& -8\\ \frac{1}{5}& 5-10-5-30& 0\end{array}$

The above table shows that $\frac{1}{5}$ is the real zero of $f\left(x\right)$.

As $f\left(\frac{1}{5}\right)=0$, then $\frac{1}{5}$ is zero. Therefore, $x-\frac{1}{5}$ is a factor of $f\left(x\right)$.

By using third row in the above table, the depressed equation is $f_1\left(x\right)=5x^3-10x^2-5x-30$.

By using the Descartes Rule of signs to find the real zeros,

$f_1\left(x\right)=5x^3-10x^2-5x-30$, and $f_1\left(-x\right)=-5x^3-10x^2+5x-30$.

The number of positive real zeros of $f_1\left(x\right)$ are $1$, since changes in the sign of the coefficient of $f_1\left(x\right)$ are $1$.

The number of negative real zeros of $f_1\left(x\right)$ are $2$, since changes in the sign of the coefficient of $f_1\left(-x\right)$ are $2$.

By using the Rational zeros Theorem to find the potential rational zeros of

$f_1\left(x\right)=5x^3-10x^2-5x-30$,

The factor of $30$; all possible value of $p$ are $\pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 10,\pm 15$.

The factor of $5$; all possible value of $q$ are $\pm 1,\pm 5$.

The potential rational zeros are $\pm 1,\pm \frac{1}{5},\pm 2,\pm \frac{2}{5}\pm 3,\pm \frac{3}{5}\pm 5,\pm 6,\pm \frac{6}{5}\pm 10,\pm 15$.

$\begin{array}{ccc}r& coefficientofquotient& remainder\\ 3& 5510& 0\end{array}$

The above table shows $3$ is the real zero of $f\left(x\right)$.

Since $f\left(3\right)=0$, then $3$ is a zero. Therefore, $x-3$ is a factor of $f\left(x\right)$.

By using fourth row in the above table, the depressed equation is $f_1\left(x\right)=5x^2+5x+10$.

Equating with zero, $5x^2+5x+10=0$

Dividing both sides of the above equation by 5 we get,

$x^2+x+2=0$

We write the function $f\left(x\right)$ as:

$f\left(x\right)=\left(x^2+x+2\right)\left(x-3\right)\left(x+0.2\right)$

Now, solve the equation $x^2+x+2=0$ by using the formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\Rightarrow \frac{-1\pm \sqrt{1^2-4(1)(2)}}{2(1)}$

$\Rightarrow \frac{-1\pm \sqrt{1-8}}{2}$

$\Rightarrow \frac{-1\pm \sqrt{-7}}{2}$

$\Rightarrow \frac{-1\pm \sqrt{7}i}{2}$.

Thus, the complex zeros of $f\left(x\right)=5x^4-9x^3-7x^2-31x-6$ are $\left\{-\frac{1}{2}+\frac{\sqrt{7}}{2}i,-\frac{1}{2}-\frac{\sqrt{7}}{2}i,-\frac{1}{5},3\right\}$.