Question

find all real zeros of the function.
$f(x)=4x(x - 16)^2(x - 6)^2$
if there is more than one answer, separate them with commas.
zero(s):

Explanation:

Step1: Set the function equal to zero

To find the real zeros of the function \( f(x) = 4x(x - 16)^2(x - 6)^2 \), we set \( f(x)=0 \). So we have the equation:
$$4x(x - 16)^2(x - 6)^2 = 0$$

Step2: Apply the zero - product property

The zero - product property states that if \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \) (or both). Since \( 4
eq0 \), we can consider the factors \( x \), \( (x - 16)^2 \) and \( (x - 6)^2 \) separately.

• For the factor \( x \):

Set \( x=0 \). When \( x = 0 \), the equation \( 4x(x - 16)^2(x - 6)^2=0 \) is satisfied because \( 4\times0\times(0 - 16)^2\times(0 - 6)^2=0 \).

• For the factor \( (x - 16)^2 \):

Set \( (x - 16)^2=0 \). Taking the square root of both sides, we get \( x-16 = 0 \), so \( x = 16 \).

• For the factor \( (x - 6)^2 \):

Set \( (x - 6)^2=0 \). Taking the square root of both sides, we get \( x - 6=0 \), so \( x = 6 \).

Answer:

0, 6, 16