Chapter 5.8, Problem 37PPS

To determine

To calculate: The zeros of the function $f\left(x\right)=48x^4-52x^3+13x-3$ .

Answer to Problem 37PPS

The zeros of the function $f\left(x\right)=48x^4-52x^3+13x-3$ are $x=-\frac{1}{2},x=\frac{1}{2},x=\frac{1}{3}\text{ and }x=\frac{3}{4}$ .

Explanation of Solution

Given information:

The function $f\left(x\right)=48x^4-52x^3+13x-3$ .

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 $f\left(x\right)=48x^4-52x^3+13x-3$ .

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 $f\left(x\right)=48x^4-52x^3+13x-3$

There are 3 or 1 positive real zeros.

Now,

  $f\left(-x\right)=48x^4+52x^3-13x-3$

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 $f\left(-x\right)=48x^4+52x^3-13x-3$ .

There is 1 negative real zero.

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

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

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 48 and constant term is 3 Therefore, p is a factor of 48 and q is a factor of 3.

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

  $\frac{p}{q}=\pm 1,\pm 3,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{1}{3},\pm \frac{1}{4},\pm \frac{3}{4},\pm \frac{1}{6},\pm \frac{1}{8},\pm \frac{3}{8},\pm \frac{1}{12},\frac{1}{16},\pm \frac{3}{16},\pm \frac{1}{24},\pm \frac{1}{48}$

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

  $\begin{array}{cccccc}x& 48& -52& 0& 13& -3\\ -1& 48& -100& 100& -87& 84\\ -3& 48& -196& 588& -1751& 5250\\ -\frac{1}{2}& 48& -38& 19& -3& 0\\ -\frac{1}{3}& 48& -\frac{68}{3}& \frac{68}{9}& \frac{49}{27}& -\frac{130}{27}\end{array}$

As observed one zero is resulted at $x=-\frac{1}{2}$ .

Now, the depressed polynomial is obtained is $24x^3-38x^2+19x-3$ . Again use synthetic substitution to factor the polynomial.

  $\begin{array}{ccccc}x& 24& -38& 19& -3\\ 1& 24& -14& 5& 2\\ \frac{1}{2}& 12& -13& 3& 0\\ \frac{1}{3}& 8& -10& 3& 0\\ \frac{1}{4}& 6& -8& \frac{11}{4}& -\frac{1}{4}\\ \frac{3}{4}& 6& -5& 1& 0\end{array}$

The zeros are obtained at $x=\frac{1}{2},x=\frac{1}{3}\text{ and }x=\frac{3}{4}$ .

Thus, the zeros of the function $f\left(x\right)=48x^4-52x^3+13x-3$ are $x=-\frac{1}{2},x=\frac{1}{2},x=\frac{1}{3}\text{ and }x=\frac{3}{4}$ .