Chapter 4, Problem 15RE

To determine

To calculate: The potential rational zeros of $f(x)=12x^8-x^7+6x^4-x^3+x-3$ .

Answer to Problem 15RE

There are 18 potential rational zeros of f .

Explanation of Solution

Given information:

The polynomial $f(x)=12x^8-x^7+6x^4-x^3+x-3$ .

Formula used:

Rational Zero Theorem:

The theorem states that if the coefficients of the polynomial $f(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{.......}+a_{1}x+a_0$ are integers, then rational zero of the polynomial can be expressed in the form $\frac{p}{q}$ where p is the factor of constant term $a_0$ and q is a factor of the leading coefficient $a_n$ .

Calculation:

Consider ,

  $f(x)=12x^8-x^7+6x^4-x^3+x-3$

where

  $\begin{array}{l}{a}_0=-3\\ a_1=1\\ a_2=0\\ a_3=-1\\ a_4=6\\ a_5=0\\ a_6=0\\ a_7=-1\\ a_8=12\end{array}$

Let coefficient of the constant term be $p=-3$ and coefficient of the leading term be $q=12$ .

  $=p=\pm 1,\pm 3$ (Factors of the constant term )

  $=q=\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12$ (Factors of the leading term )

Now, the possible ratios are :-

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

If f has rational zeros, it will one of ratio mentioned in the list, which contain 18 possibilities.