To find the zeros of the function [tex]\( y = (x - 4)(x^2 - 12x + 36) \)[/tex], we need to determine the values of [tex]\( x \)[/tex] that make [tex]\( y = 0 \)[/tex]. The function [tex]\( y \)[/tex] is already factored as the product of two expressions: [tex]\( x - 4 \)[/tex] and [tex]\( x^2 - 12x + 36 \)[/tex]. We will solve for [tex]\( x \)[/tex] by setting each factor to zero.
1. Set [tex]\( x - 4 = 0 \)[/tex]:
[tex]\[
x - 4 = 0
\][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[
x = 4
\][/tex]
2. Set [tex]\( x^2 - 12x + 36 = 0 \)[/tex]:
[tex]\[
x^2 - 12x + 36 = 0
\][/tex]
Notice that [tex]\( x^2 - 12x + 36 \)[/tex] is a perfect square trinomial. It can be factored as:
[tex]\[
x^2 - 12x + 36 = (x - 6)^2
\][/tex]
Therefore, we have:
[tex]\[
(x - 6)^2 = 0
\][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[
x - 6 = 0
\][/tex]
[tex]\[
x = 6
\][/tex]
Thus, the zeros of the function [tex]\( y = (x - 4)(x^2 - 12x + 36) \)[/tex] are [tex]\( x = 4 \)[/tex] and [tex]\( x = 6 \)[/tex].
The correct answer is:
B. 4 and 6
