Question

what are the zeros of the function $y = (x - 4)(x^2 - 12x + 36)$?

a. -6, 4, and 6

b. 4 and 6

c. -6 and -4

d. 0, 4, and 6

Explanation:

Step1: Find zero from first factor

Set \(x - 4 = 0\), solve for \(x\): \(x = 4\).

Step2: Factor the quadratic

Factor \(x^2 - 12x + 36\). It's a perfect square: \((x - 6)^2\).

Step3: Find zero from quadratic factor

Set \((x - 6)^2 = 0\), solve for \(x\): \(x = 6\) (double root).

Step4: Combine zeros

Zeros are \(4\) and \(6\) (with \(6\) repeated, but as a zero, we list it once in the context of options).

Answer:

B. 4 and 6