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What are the zeros of the function [tex]$y=(x-4)\left(x^2-12x+36\right)$[/tex]?

A. [tex]-6, 4[/tex], and [tex]6[/tex]
B. [tex]4[/tex] and [tex]6[/tex]
C. [tex]-6[/tex] and [tex]-4[/tex]
D. [tex]0, 4[/tex], and [tex]6[/tex]

Sagot :

GinnyAnswer
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
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