Find an nth degree polynomial. N=4, 2i and 3i are zeros, f (-1)=50
since complex roots come in conjugate pairs,
f(x) = a(x-2i)(x+2i)(x-3i)(x+3i)
= a(x^2+4)(x^2+9)
Now use the given point to find a.
To find an nth degree polynomial, we need n+1 values or information. Since the degree of the polynomial is given as 4, we need five pieces of information.
First, we are given that 2i and 3i are zeros, which means that (x - 2i) and (x - 3i) are factors of the polynomial. Since complex zeros occur in conjugate pairs, we also have (x + 2i) and (x + 3i) as factors.
Now, let's substitute (-1) for x in the polynomial and set it equal to 50, as given by f(-1) = 50.
Since (-1 + 2i) and (-1 + 3i) are not zero, we can multiply them out to determine the corresponding factors:
(x - 2i) * (x + 2i) = x^2 - (2i)^2 = x^2 + 4
(x - 3i) * (x + 3i) = x^2 - (3i)^2 = x^2 + 9
Now we can write the polynomial equation based on the given information:
P(x) = a(x - 2i)(x + 2i)(x - 3i)(x + 3i) = a(x^2 + 4)(x^2 + 9)
To determine the value of 'a', we substitute (-1) for 'x' in the polynomial equation and equate it to 50, as given:
P(-1) = a((-1)^2 + 4)((-1)^2 + 9) = 50
Simplifying this equation:
a(1 + 4)(1 + 9) = 50
a(5)(10) = 50
50a = 50
a = 1
Therefore, the nth degree polynomial is:
P(x) = (x^2 + 4)(x^2 + 9)
This is a fourth-degree polynomial with zeros at 2i and 3i, and satisfies the condition f(-1) = 50.
To find an nth degree polynomial with 2i and 3i as zeros, we need to consider their conjugates as well. Thus, the zeros of the polynomial will be 2i, -2i, 3i, and -3i.
Since complex roots come in conjugate pairs, we know that if 2i is a zero, then -2i must also be a zero, and if 3i is a zero, then -3i must also be a zero.
Therefore, the factors of the polynomial are (x - 2i)(x + 2i)(x - 3i)(x + 3i).
To find the specific polynomial equation, we can multiply these factors together:
(x - 2i)(x + 2i) = x^2 - (2i)^2 = x^2 + 4
(x - 3i)(x + 3i) = x^2 - (3i)^2 = x^2 + 9
Now we can multiply the two resulting expressions together:
(x^2 + 4)(x^2 + 9) = x^4 + 9x^2 + 4x^2 + 36 = x^4 + 13x^2 + 36
So, the nth degree polynomial with zeros 2i and 3i is f(x) = x^4 + 13x^2 + 36.
Given f(-1) = 50, we can substitute x = -1 into the polynomial:
f(-1) = (-1)^4 + 13(-1)^2 + 36 = 1 + 13 + 36 = 50
Therefore, the polynomial f(x) = x^4 + 13x^2 + 36 satisfies the given conditions.