Final answer:
The equation of a real function with zeroes at x = 4 and x = -1 + 2i (assuming typo correction) is f(x) = (x - 4)(x^2 + 2x - 3). To find a real polynomial, complex roots always come in conjugate pairs.
Step-by-step explanation:
To write an equation of a function with given zeroes, you essentially want to find a polynomial that has these values as its roots. The given zeroes are x = 4 and x = -1 + 26. However, the latter seems to be a typo or a misunderstanding because 26 is a real number, and adding it to -1 would not produce a usually intended complex number (which would be excluded from a real function).
Assuming the second zero should read x = -1 + 2i, which is a complex number, we know that real polynomials have complex roots that occur in conjugate pairs. Therefore, if you have a complex zero of -1 + 2i, there must also be a corresponding zero at -1 - 2i for the function to be a real polynomial.
The polynomial with real coefficients that has these zeros can thus be written as:
f(x) = (x - 4)(x - (-1 + 2i))(x - (-1 - 2i))
Expanding the factors gives us the equation of the polynomial:
f(x) = (x - 4)((x + 1) - 2i)((x + 1) + 2i)
f(x) = (x - 4)((x + 1)^2 + (2i)^2)
f(x) = (x - 4)(x^2 + 2x + 1 - 4)
f(x) = (x - 4)(x^2 + 2x - 3)