Chapter EP, Problem 6.8.10EP

To determine

To calculate: The zeros of the function $f\left(x\right)=x^4+8x^2-9$ .

Answer to Problem 6.8.10EP

The zeros of the function $f\left(x\right)=x^4+8x^2-9$ are $1,-1,3i,-3i$ .

Explanation of Solution

Given information:

The function $f\left(x\right)=x^4+8x^2-9$ .

Formula used:

A polynomial of n degree has n zeros, which can be either real or imaginary.

Descartes’ rule of signs states that consider a polynomial $P\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ with real coefficients then, the number of times the sign between coefficients changes is equal to count of positive real zeroes of $P\left(x\right)$ or is less than this by an even number, the number of times the sign between coefficients changes is equal to count of negative real zeroes of $P\left(-x\right)$ or is less than this by an even number.

Calculation:

Consider the function $f\left(x\right)=x^4+8x^2-9$ .

Observe that degree of polynomial is 4, so the functions has 4 zeros which can be either real or imaginary.

Descartes’ rule of signs states that consider a polynomial $P\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ with real coefficients then, the number of times the sign between coefficients changes is equal to count of positive real zeroes of $P\left(x\right)$ or is less than this by an even number.

So, count the number of times the sign changes between the coefficients of $f\left(x\right)=x^4+8x^2-9$

There is 1 positive real zero.

Now,

  $\begin{array}{c}f\left(-x\right)={\left(-x\right)}^4+8{\left(-x\right)}^2-9\\ f\left(-x\right)=x^4+8x^2-9\end{array}$

Descartes’ rule of signs states that consider a polynomial $P\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ with real coefficients then, the number of times the sign between coefficients changes is equal to count of negative real zeroes of $P\left(-x\right)$ or is less than this by an even number.

So, count the number of times the sign changes between the coefficients of $f\left(-x\right)=x^4+8x^2-9$ .

There is 1 negative real zero.

Next, construct a table with possible combinations of real and imaginary zeros.

  $\begin{array}{cccc}\begin{array}{c}\text{Number of Positive}\\ \text{Real zeros}\end{array}& \begin{array}{c}\text{Number of Negative}\\ \text{Real zeros}\end{array}& \begin{array}{c}\text{Number of}\\ \text{Imaginary zeros}\end{array}& \begin{array}{c}\text{Total Number}\\ \text{of zeros}\end{array}\\ 1& 1& 2& 4\end{array}$

Recall that the Rational zero theorem states that provided a polynomial $P\left(x\right)$ with integral coefficients then every rational zero is of the form $\frac{p}{q}$ that is a rational number in simplest form. Here, p is a factor of the constant term and q is a factor of leading coefficient.

For the provided function leading coefficient is 1 and constant term is 9 Therefore, p is a factor of 9 and q is a factor of 1.

The possible combinations of $\frac{p}{q}$ in simplest form are,

  $\frac{p}{q}=\pm 1,\pm 3,\pm 9$

Next, construct a table with help of synthetic substitution to compute the value of $f\left(x\right)$ for real values of $x$ .

  $\begin{array}{cccccc}\frac{p}{q}& 1& 0& 8& 0& -9\\ -1& 1& -1& 9& -9& 0\\ 1& 1& 1& 9& 9& 0\\ 3& 1& 3& 17& 51& 144\\ -3& 1& -3& 5& -51& 144\end{array}$

As observed one zero is resulted at $x=-1$ . Since, only one real negative zero is obtained so use the depressed polynomial to obtain remaining zeroes.

Now, the depressed polynomial is obtained is $f\left(x\right)=x^3-x^2+9x-9$ .

Group the terms together and factor the polynomial.

  $\begin{array}{c}{x}^3-x^2+9x-9=0\\ x^2\left(x-1\right)+9\left(x-1\right)=0\\ \left(x-1\right)\left(x^2+9\right)=0\end{array}$

Apply the zero product property.

Either $\left(x-1\right)=0\text{ or }\left(x^2+9\right)=0$

That is, $x=1,x=\pm i3$ .

Thus, the zeros of the function $f\left(x\right)=x^4+8x^2-9$ are $1,-1,3i,-3i$ .