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What is a quartic function with only the two real zeroes given?

Zeroes: \(x = 5\) and \(x = 1\)

A. \(9 + x^9 - 2x^9 + 5x^9 = 0\)

B. \(9 - x^9 + 2x^9 - x^9 + x = 0\)

C. \(9 + x^9 - 2x^9 + x^9 - x = 0\)

D. \(9 + x^9 - 2x^9 + x^9 - x = 0\)

Answer :

Final answer:

A quartic function with the given real zeroes is f(x) = x^2 - 6x + 5.


Explanation:

A quartic function is a function of the form f(x) = ax^4 + bx^3 + cx^2 + dx + e where a, b, c, d, and e are constants.

Given that the real zeroes of the function are x = 5 and x = 1, we can write the factors of the function as (x - 5) and (x - 1). To find the quartic function, we can multiply these factors together:

f(x) = (x - 5)(x - 1)

Expanding this using the distributive property, we get:

f(x) = x^2 - 6x + 5


Learn more about Quartic Functions here:

https://brainly.com/question/30180400


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