To find a fourth-degree polynomial function with real coefficients that satisfies the given conditions, we need to consider the zeros and the y-intercept.
Since -3, 1/5, and 4+i are zeros of the polynomial, their conjugates (-3, 1/5, and 4-i) must also be zeros. This is because polynomials with real coefficients have complex zeros in conjugate pairs.
Now, let's start by writing the equation in factored form:
f(x) = a(x - (-3))(x - 1/5)(x - (4+i))(x - (4-i))
To simplify, let's multiply the terms:
f(x) = a(x + 3)(x - 1/5)(x - 4 - i)(x - 4 + i)
Multiplying the last two factors:
f(x) = a(x + 3)(x - 1/5)[(x - 4) - i][(x - 4) + i]
Applying the difference of squares formula:
f(x) = a(x + 3)(x - 1/5)[(x - 4)^2 - (i)^2]
Since (i)^2 is equal to -1:
f(x) = a(x + 3)(x - 1/5)[(x - 4)^2 + 1]
Expanding the squared term:
f(x) = a(x + 3)(x - 1/5)(x^2 - 8x + 16 + 1)
Simplifying further:
f(x) = a(x + 3)(x - 1/5)(x^2 - 8x + 17)
To find the value of "a" and satisfy the given y-intercept of (0,5), we substitute x=0 and y=5 into the equation:
5 = a(0 + 3)(0 - 1/5)(0^2 - 8(0) + 17)
5 = a(3)(-1/5)(17)
5 = a(-3)(17/5)
25 = -51a
a = -25/51
Therefore, the final fourth-degree polynomial function with real coefficients that satisfies the given conditions is:
f(x) = (-25/51)(x + 3)(x - 1/5)(x^2 - 8x + 17)