Chapter 5.8, Problem 35PPS

To determine

To calculate: The zeros of the function $f\left(x\right)=9x^5-94x^3+27x^2+40x-12$ .

Answer to Problem 35PPS

The zeros of the function $f\left(x\right)=9x^5-94x^3+27x^2+40x-12$ are $3,\pm \frac{2}{3}\text{ and }\frac{-3\pm \sqrt{13}}{2}$ .

Explanation of Solution

Given information:

The function $f\left(x\right)=9x^5-94x^3+27x^2+40x-12$ .

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 $f\left(x\right)=9x^5-94x^3+27x^2+40x-12$ .

Observe that degree of polynomial is 5, so the functions has 5 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 $f\left(x\right)=9x^5-94x^3+27x^2+40x-12$

There are 3 or 1 positive real zeros.

Now,

  $f\left(-x\right)=-9x^5+94x^3+27x^2-40x-12$

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 $f\left(-x\right)=-9x^5+94x^3+27x^2-40x-12$ .

There are 2 or 0 negative real zero.

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

  $\begin{array}{cccc}\begin{array}{c}\text{Number of Positive}\\ \text{Real zeros}\end{array}& \begin{array}{c}\text{Number of Negative}\\ \text{Real zeros}\end{array}& \begin{array}{c}\text{Number of}\\ \text{Imaginary zeros}\end{array}& \begin{array}{c}\text{Total Number}\\ \text{of zeros}\end{array}\\ 3& 0& 2& 3+0+2=5\\ 3& 2& 0& 3+2+0=5\\ 1& 0& 4& 1+0+4=5\\ 1& 2& 2& 1+2+2=5\end{array}$

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 9 and constant term is 12. Therefore, p is a factor of 12 and q is a factor of 9.

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

  $\frac{p}{q}=\pm 1,\pm \frac{1}{3},\pm \frac{1}{9},\pm 2,\pm \frac{2}{3},\pm \frac{2}{9},\pm 4,\pm \frac{4}{3},\pm \frac{4}{9},\pm 6,\pm 12$

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

  $\begin{array}{ccccccc}x& 9& 0& -94& 27& 40& -12\\ 3& 9& 27& -13& -12& 4& 0\\ -3& 9& -27& -13& 66& -158& 462\\ \frac{2}{3}& 9& 6& -90& -33& 18& 0\\ -\frac{2}{3}& 9& -6& -90& 87& -18& 0\end{array}$

As observed three zeros are resulted at $x=2,x=\frac{2}{3}\text{ and }x=-\frac{2}{3}$ .

So, factor of the polynomial is,

  $\begin{array}{c}\left(x-2\right)\left(x-\frac{2}{3}\right)\left(x+\frac{2}{3}\right)=\left(x-2\right)\left(x^2-\frac{4}{9}\right)\\ =x^3-2x^2-\frac{4}{9}x+\frac{8}{9}\end{array}$ .

Now, divide the polynomial $f\left(x\right)=9x^5-94x^3+27x^2+40x-12$ by $x^3-2x^2-\frac{4}{9}x+\frac{8}{9}$ .

  $9x^5-94x^3+27x^2+40x-12=\left(x^3-2x^2-\frac{4}{9}x+\frac{8}{9}\right)\left(9x^2+27x-9\right)$

Now, the depressed polynomial is obtained is $\left(9x^2+27x-9\right)$ .

Factor the polynomial.

  $\begin{array}{c}{x}^2+3x-1=0\\ x=\frac{-3\pm \sqrt{13}}{2}\end{array}$

That is, $\frac{-3+\sqrt{13}}{2}\text{ or }\frac{-3-\sqrt{13}}{2}$ .

Thus, the zeros of the function $f\left(x\right)=9x^5-94x^3+27x^2+40x-12$ are $3,\pm \frac{2}{3}\text{ and }\frac{-3\pm \sqrt{13}}{2}$ .