Chapter 2.6, Problem 38E

Solution

To find: a polynomial function $f$ .

A polynomial $f\left(x\right)=4x+3$ .

Given information: 

It is a leading coefficient of $4$ and has $12$ possible rational zeros.

Formula used:

The Rational Zero Theorem:

If $f\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ consists of integer coefficients, then the shape of each rational zero of $f$ is as follows:

  $\frac{p}{q}=\frac{\text{ factor of constant term }{a}_0}{\text{ factor of leading coefficient }{a}_n}$

Calculation:

Therefore, all that is left to do is identify the proper constant term.

Start by examining the data that the leading coefficient of $4$ provides.

Factors of $4$ :

  $\pm 1,\pm 2,\pm 4$

Consequently, all potential rational zeros have the form

  $\frac{p}{1},\frac{p}{2},\frac{p}{4}$

where $p$ is a factor of the constant term.

Therefore, each whole numbered constant term component provides us with $6$ additional potential zeros (remember around $\pm$ !)

In light of this, the constant term must be a number with only two elements that are whole numbers, or a prime number.

For example, as a constant term, consider $3$ , Use the simplest function; we can change the other coefficients and the degree of the function.

  $f\left(x\right)=4x+3$

Factors of $3$ :

  $\pm 1,\pm 3$

As a result, all conceivable zeros are (enter all possible permutations of $4$ and $3$ ):

  $\pm \frac{1}{1},\pm \frac{3}{1},\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{1}{4},\pm \frac{3}{4}$

So, there are indeed $12$ possible rational zeros (again, remember that $\pm$ gives two

numbers).