Chapter EP, Problem 6.8.6EP

To determine

To calculate: The rational zeros of the function $f\left(x\right)=20x^4-16x^3+11x^2-12x-3$ .

Answer to Problem 6.8.6EP

The rational zeros of the function $f\left(x\right)=20x^4-16x^3+11x^2-12x-3$ are $1\text{ and }-\frac{1}{5}$ .

Explanation of Solution

Given information:

The function $f\left(x\right)=20x^4-16x^3+11x^2-12x-3$ .

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 $f\left(x\right)=20x^4-16x^3+11x^2-12x-3$ .

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 $f\left(x\right)=20x^4-16x^3+11x^2-12x-3$

There are 3 sign changes, so there are 3 or 1 positive real zeros.

Now,

  $\begin{array}{c}f\left(-x\right)=20{\left(-x\right)}^4-16{\left(-x\right)}^3+11{\left(-x\right)}^2-12\left(-x\right)-3\\ f\left(-x\right)=20x^4+16x^3+11x^2+12x-3\end{array}$

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 $f\left(-x\right)=20x^4+16x^3+11x^2+12x-3$ .

There is 1 sign change, so there is 1 negative real zero.

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

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

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 20 and constant term is $-3$ . Therefore, p is a factor of $3$ and q is a factor of 20.

Factors of $3$ are $\pm 1\text{ and }\pm \text{3}$ .

Factors of 20 are $\pm 1,\pm 2,\pm 4,\pm 5,\pm 10\text{ and }\pm 20$

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

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

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

  $\begin{array}{cccccc}\frac{p}{q}& 20& -16& 11& -12& -3\\ -3& 20& -76& 239& -729& 2184\\ -1& 20& -36& 47& -59& 56\\ 1& 20& 4& 15& 3& 0\\ 3& 20& 44& 143& 417& 1248\end{array}$

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

Now, the depressed polynomial is obtained is $f\left(x\right)=20x^3+4x^2+15x+3$ .

Group the terms together and factor the polynomial.

  $\begin{array}{c}20x^3+4x^2+15x+3=0\\ 4x^2\left(5x+1\right)+3\left(5x+1\right)=0\\ \left(5x+1\right)\left(4x^2+3\right)=0\end{array}$

Apply the zero product property.

Either $\left(5x+1\right)=0\text{ or }\left(4x^2+3\right)=0$

That is, $x=-\frac{1}{5},x=\pm i\frac{\sqrt{3}}{2}$ .

Thus, the rational zeros of the function $f\left(x\right)=20x^4-16x^3+11x^2-12x-3$ are $1\text{ and }-\frac{1}{5}$ .