Question

find all real zeros of the function.
$f(x) = 3(x - 5)(x^2 - 36)(x + 5)^2$
if there is more than one answer, separate them with commas.
zero(s):

Explanation:

Step1: Set each factor to zero

To find the real zeros of the function \( f(x) = 3(x - 5)(x^2 - 36)(x + 5)^2 \), we set each factor equal to zero.

• For \( x - 5 = 0 \), we solve for \( x \):

\( x - 5 = 0 \implies x = 5 \)

• For \( x^2 - 36 = 0 \), we factor the difference of squares:

\( x^2 - 36 = (x - 6)(x + 6) = 0 \)
Setting each factor equal to zero gives \( x - 6 = 0 \implies x = 6 \) and \( x + 6 = 0 \implies x = -6 \)

• For \( (x + 5)^2 = 0 \), we take the square root of both sides:

\( (x + 5)^2 = 0 \implies x + 5 = 0 \implies x = -5 \) (this root has multiplicity 2, but we still list it once as a real zero)

Step2: Combine the zeros

The real zeros are the solutions we found: \( 5, -6, 6, -5 \)

Answer:

-5, 5, -6, 6