Chapter 5.8, Problem 33PPS

To determine

To calculate: The zeros of the function $f\left(x\right)=6x^4+22x^3+11x^2-38x-40$ .

Answer to Problem 33PPS

The zeros of the function $f\left(x\right)=6x^4+22x^3+11x^2-38x-40$ are $-2,\frac{4}{3},\frac{-3+i}{2}\text{ and }\frac{-3-i}{2}$ .

Explanation of Solution

Given information:

The function $f\left(x\right)=6x^4+22x^3+11x^2-38x-40$ .

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 $f\left(x\right)=6x^4+22x^3+11x^2-38x-40$ .

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 $f\left(x\right)=6x^4+22x^3+11x^2-38x-40$

There is 1 positive real zero.

Now,

  $f\left(-x\right)=6x^4-22x^3+11x^2+38x-40$

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 $f\left(-x\right)=6x^4-22x^3+11x^2+38x-40$ .

There are 3 or 1 negative real zeros.

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

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

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 6 and constant term is 40 Therefore, p is a factor of 40 and q is a factor of 6.

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

  $\frac{p}{q}=\pm 1,\pm 2,\pm 4,\pm 5,\pm 8,\pm 10,\pm 20,\pm 40,\pm \frac{1}{2},\pm \frac{5}{2},\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{4}{3},\pm \frac{5}{3},\pm \frac{8}{3},\pm \frac{10}{3},\pm \frac{20}{3},\pm \frac{40}{3},\pm \frac{1}{6},\pm \frac{5}{6}$

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

  $\begin{array}{cccccc}x& 6& 22& 11& -38& -40\\ 1& 6& 28& 39& 1& -39\\ -1& 6& 16& -5& -33& -7\\ 2& 6& 34& 79& 120& 200\\ -2& 6& 10& -9& -20& 0\\ \frac{1}{2}& 6& \frac{25}{2}& \frac{47}{4}& -\frac{105}{8}& -\frac{425}{8}\\ -\frac{1}{2}& 6& \frac{19}{2}& \frac{3}{4}& -\frac{155}{8}& -\frac{165}{8}\\ \frac{4}{3}& 6& 10& 17& 10& 0\\ -\frac{4}{3}& 6& \frac{14}{3}& -\frac{23}{9}& -\frac{250}{27}& -\frac{80}{27}\end{array}$

As observed two zeros are resulted at $x=-2\text{ and }x=\frac{4}{3}$ . One of the factors of the polynomial is

  $\left(x+2\right)\left(3x-4\right)=3x^2+2x-8$

Now, divide the polynomial $f\left(x\right)=6x^4+22x^3+11x^2-38x-40$ by $\left(3x^2+2x-8\right)$ to obtain remaining factors.

  $3x^2+2x-8\begin{array}{c}\hfill 2x^2+6x+5\\ \hfill \overline{)\begin{array}{l}\begin{array}{ccccccccc}6x^4& +& 22x^3& +& 11x^2& -& 38x& -& 40\end{array}\\ \begin{array}{cccccc}-6x^4& -& 4x^3& +& 16x^2& \end{array}\\ \begin{array}{cccccccc}& & & 18x^3& +27x^2& -& 38x& \end{array}\\ \begin{array}{cccccccc}& & & -18x^3& -12x^2& +& 48x& \end{array}\\ \begin{array}{ccccccccccc}& & & & & & 15x^2& +& 10x& -& 40\end{array}\\ \begin{array}{ccccccccccc}& & & & & & -15x^2& -10x& +& 40& \end{array}\\ \begin{array}{ccccccccccc}& & & & & & & & & & 0\end{array}\end{array}}\end{array}$

Now, the depressed polynomial is obtained is $2x^2+6x+5$ .

Factor the polynomial.

  $\begin{array}{c}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{-6\pm \sqrt{6^2-4\left(2\right)\left(5\right)}}{2\left(2\right)}\\ x=\frac{-6\pm \sqrt{-4}}{4}\\ x=\frac{-6\pm 2i}{4}\end{array}$

Either $x=\frac{-6+2i}{4}\text{ or }x=\frac{-6-2i}{4}$

That is, $x=\frac{-3+i}{2}\text{ or }x=\frac{-3-i}{2}$ .

Thus, the zeros of the function $f\left(x\right)=6x^4+22x^3+11x^2-38x-40$ are $-2,\frac{4}{3},\frac{-3+i}{2}\text{ and }\frac{-3-i}{2}$ .