Question

use factoring to find the real zeros of the function.
$g(x) = x^2 - 289$

select the correct choice and, if necessary, fill in the answer box to complete your choice.

a. the real zero(s) of the function is/are $x = $
(type an integer or a simplified fraction. use a comma to separate answers as needed.)

b. there are no real zeros.

Explanation:

Step1: Recognize the difference of squares

The function \( g(x) = x^2 - 289 \) is a difference of squares, since \( 289 = 17^2 \). The formula for factoring a difference of squares is \( a^2 - b^2 = (a - b)(a + b) \).
So, \( x^2 - 289 = x^2 - 17^2 = (x - 17)(x + 17) \).

Step2: Find the zeros

To find the real zeros, we set \( g(x) = 0 \):
\( (x - 17)(x + 17) = 0 \)
Using the zero - product property, if \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \).
Case 1: \( x - 17 = 0 \), then \( x = 17 \).
Case 2: \( x + 17 = 0 \), then \( x=- 17 \).

Answer:

The real zero(s) of the function is/are \( x = - 17,17 \)