Question

use the following function to answer parts a through c. f(x)=5x^3 + 27x^2 + 11x + 5
a. list all possible rational zeros.
1, - 1,5, - 5,\frac{1}{5},-\frac{1}{5}
(type an integer or a simplified fraction. use a comma to separate answers as needed. type each answer only once.)
b. use synthetic division to test the possible rational zeros and find an actual zero.
one rational zero of the given function is .
(simplify your answer.)

Explanation:

Step1: Recall Rational Zero Theorem

The Rational Zero Theorem states that if a polynomial function \(f(x)=a_nx^n + a_{n - 1}x^{n-1}+\cdots+a_1x + a_0\) has integer coefficients, then the possible rational zeros are of the form \(\frac{p}{q}\), where \(p\) is a factor of the constant term \(a_0\) and \(q\) is a factor of the leading - coefficient \(a_n\). For \(f(x)=5x^{3}+27x^{2}+11x + 5\), \(a_0 = 5\) and \(a_n=5\).

Step2: Find factors of \(a_0\) and \(a_n\)

The factors of \(a_0 = 5\) are \(p=\pm1,\pm5\), and the factors of \(a_n = 5\) are \(q=\pm1,\pm5\).

Step3: Calculate possible rational zeros

The possible rational zeros \(\frac{p}{q}\) are \(\frac{\pm1}{\pm1},\frac{\pm5}{\pm1},\frac{\pm1}{\pm5},\frac{\pm5}{\pm5}\), which simplifies to \(1, - 1,5,-5,\frac{1}{5},-\frac{1}{5}\).

Step4: Use synthetic division to test zeros

Let's test \(x=-5\) using synthetic division.
Set up the synthetic - division:

-5 |  5   27   11   5
    |     -25  -10 -5
    |----------------
      5    2    1   0

Since the remainder is \(0\), \(x=-5\) is a zero of the function \(f(x)\).

Answer:

a. \(1,-1,5,-5,\frac{1}{5},-\frac{1}{5}\)
b. \(-5\)