write a fourth degree polynomial function with real coefficients that has -3,1/5, and 4+i as zeros and the y intercept of (0,5)
The other root must be 4-i, so one factor is
(x-(4+i))(x-(4-i)) = ((x-4)^2 + 1^2) = x^2-8x+17
So, now we have
(x+3)(5x-1)(x^2-8x+17)
at x=0, this is (3)(-1)(17) = -51, so
y = -5/51 (x+3)(5x-1)(x^2-8x+17)
= -5/51 (5x^4 - 26x^3 - 30x^2 + 262x - 51)
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)