Chapter EP, Problem 6.8.12EP

To determine

To calculate: The zeros of the function $f\left(x\right)=4x^4+19x^2-63$ .

Answer to Problem 6.8.12EP

The zeros of the function $f\left(x\right)=4x^4+19x^2-63$ are $\frac{3}{2},-\frac{3}{2},\sqrt{7}i,-\sqrt{7}i$ .

Explanation of Solution

Given information:

The function $f\left(x\right)=4x^4+19x^2-63$ .

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 $f\left(x\right)=4x^4+19x^2-63$ .

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 $f\left(x\right)=4x^4+19x^2-63$

There is 1 positive real zero.

Now,

  $f\left(-x\right)=4x^4+19x^2-63$

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 $f\left(-x\right)=4x^4+19x^2-63$ .

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 4 and constant term is 63. Therefore, p is a factor of 63 and q is a factor of 4.

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

  $\frac{p}{q}=\pm 1,\pm \frac{3}{2},\pm \frac{7}{2}$

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

  $\begin{array}{cccccc}\frac{p}{q}& 4& 0& 19& 0& -63\\ -1& 4& -4& 23& -23& -40\\ 1& 4& 4& 23& 23& -40\\ \frac{3}{2}& 4& 6& 28& 42& 0\\ -\frac{3}{2}& 4& -6& 28& -42& 0\end{array}$

As observed one zero is resulted at $x=\frac{3}{2}$ . Since, only one real positive zero is obtained so use the depressed polynomial to obtain remaining zeroes.

Now, the depressed polynomial is obtained is $f\left(x\right)=4x^3+6x^2+28x+42$ .

Group the terms together and factor the polynomial.

  $\begin{array}{c}4x^3+6x^2+28x+42=0\\ 2x^2\left(2x+3\right)+14\left(2x+3\right)=0\\ \left(2x+3\right)\left(2x^2+14\right)=0\end{array}$

Apply the zero product property.

Either $\left(2x+3\right)=0\text{ or }\left(2x^2+14\right)=0$

That is, $x=-\frac{3}{2},x=\pm \sqrt{7}i$ .

Thus, the zeros of the function $f\left(x\right)=4x^4+19x^2-63$ are $\frac{3}{2},-\frac{3}{2},\sqrt{7}i,-\sqrt{7}i$ .