Answer:
Step-by-step explanation:
To write the equation for a cubic polynomial function whose graph has zeroes at 2, 3, and 5, we start by using the fact that these roots imply the polynomial can be written in the form:
P(x) = a(x - 2)(x - 3)(x - 5)
Here, a is a non-zero constant that can be any real number. If we assume a = 1 for simplicity, the polynomial becomes:
P(x) = (x - 2)(x - 3)(x - 5)
Now, let’s expand this polynomial:
First, expand \(x - 2)(x - 3):
(x - 2)(x - 3) = x^2 - 3x - 2x + 6 = x^2 - 5x + 6
Next, multiply this result by (x - 5):
(x^2 - 5x + 6)(x - 5)
= x^3 - 5x^2 + 6x - 5x^2 + 25x - 30
= x^3 - 10x^2 + 31x - 30
Thus, the cubic polynomial with zeroes at 2, 3, and 5 is:
P(x) = x^3 - 10x^2 + 31x - 30
Regarding multiplicity, none of the roots 2, 3, or 5 have a multiplicity greater than 1 in the current polynomial because each factor (x - 2), (x - 3), and (x - 5) appears exactly once.
If any root were to have a multiplicity greater than 1, the polynomial would no longer be cubic; instead, it would have a higher degree. For example, if the root (2) had a multiplicity of 2, the polynomial would be:
P(x) = a(x - 2)^2(x - 3)(x - 5)
This polynomial is quartic (degree 4), not cubic.
