how do I find a 4th degree polynomial that has zeros; i,3i with f(-1)=60

I know the conjucants are -i and -3i

(x-i)(x-3i) (x+i)(x=3i)

So, multiply it out to get

f(x) = a(x^2+1)(x^2+9) = a(x^4+10x^2+9)

Now, f(-1) = a(1+10+9) = 20a, so we need a=3 and thus

f(x) = 3(x^4+10x^2+9)

BTW, that's "conjugates"

To find a 4th degree polynomial with zeros i, 3i, and f(-1) = 60, we can use the fact that complex zeros occur in conjugate pairs.

Step 1: Set up the polynomial in factored form using the given zeros:
(x - i)(x - 3i)(x + i)(x + 3i)

Step 2: Simplify the factored form:
(x^2 - ix - 3ix + 3i^2)(x^2 + ix + 3ix + 3i^2)

Step 3: Simplify further:
(x^2 - 4ix - 3)(x^2 + 4ix - 3)

Step 4: Expand the simplified expression:
(x^2 - 4ix - 3)(x^2 + 4ix - 3) = x^4 + 4ix^3 - 3x^2 - 4ix^3 - 16ix^2 + 12x - 3x^2 - 12ix + 9

Step 5: Collect the like terms:
x^4 + (4ix^3 - 4ix^3) - (3x^2 + 16ix^2 + 3x^2) + 12x - 12ix + 9

Simplifying further:
x^4 - 19x^2 + 12x - 12ix + 9

Thus, a fourth-degree polynomial with zeros i, 3i, and f(-1) = 60 is:
f(x) = x^4 - 19x^2 + 12x - 12ix + 9.

To find a 4th degree polynomial with zeros i, 3i, and satisfy f(-1) = 60, we can use the fact that complex conjugate roots always come in pairs. Therefore, the zeros will be (i, -i, 3i, -3i).

We'll start by setting up the equation using the formula for a polynomial given its zeros. Let the polynomial be denoted by f(x):

f(x) = (x - i)(x + i)(x - 3i)(x + 3i)

Simplifying this expression, we get:

f(x) = (x^2 + 1)(x^2 + 9)

Now, we substitute f(-1) = 60 into the equation:

f(-1) = (-1^2 + 1)(-1^2 + 9) = (1)(8) = 8

We can see that f(-1) = 8, not 60, so the polynomial we obtained is not correct.

To satisfy f(-1) = 60, we need to modify the polynomial by multiplying it with a constant. Let's call this constant 'a':

f(x) = a(x - i)(x + i)(x - 3i)(x + 3i)

Now, let's use the given information that f(-1) = 60 and substitute it into the equation:

60 = a(-1 - i)(-1 + i)(-1 - 3i)(-1 + 3i)

Expand and simplify the expression:

60 = a(2)(10)(10) = 200a

Divide by 200 to solve for a:

a = 60/200 = 0.3

Therefore, the 4th degree polynomial that satisfies f(-1) = 60 and has zeros i and 3i is:

f(x) = 0.3(x - i)(x + i)(x - 3i)(x + 3i)