Can you please help with this one.

Find an​ nth-degree polynomial function with real coefficients satisfying the given conditions.

n=4
2 i and 4 i are zeros;

f(-1)=85
f(x)=
​(Type an expression using x as the variable. Simplify your​ answer.)

if it has real coefficients, then its complex roots come in conjugate pairs, meaning

f(x) = a(x-2i)(x+2i)(x-4i)(x+4i)
= a(x^2+4)(x^2+16)

Now just find a that makes f(-1) = 85

for n repeated independent trials, with a constant probability of success p for all trials, find the probability of exactly x success in each of them

a) n=5, p=1/3, x=4
b) n=30, p=0.6, x=22

Well, finding polynomial functions can be quite a hair-raising task! But don't worry, I'm here to clown around and lend you a hand.

Since 2i and 4i are zeros of the polynomial function, we know that their conjugates must also be zeros. So, the zeros of the polynomial function are: 2i, 4i, -2i, and -4i.

Now, remember that for a polynomial with real coefficients, the complex zeros must occur in conjugate pairs. That means if 2i is a zero, then -2i must also be a zero, and if 4i is a zero, then -4i must also be a zero.

With that in mind, we can construct our polynomial function using these zeros. The general form of a fourth-degree polynomial is:

f(x) = a(x - r1)(x - r2)(x - r3)(x - r4)

where r1, r2, r3, and r4 are the zeros and a is a constant.

Plugging in our zeros, we get:

f(x) = a(x - 2i)(x + 2i)(x - 4i)(x + 4i)

Now, let's simplify this expression. The complex conjugates will give us a difference of squares:

(x - 2i)(x + 2i) = x^2 - (2i)^2 = x^2 - 4

(x - 4i)(x + 4i) = x^2 - (4i)^2 = x^2 + 16

Substituting these simplified expressions back into f(x), we get:

f(x) = a(x^2 - 4)(x^2 + 16)

Now, we just need to find the value of a. To do that, we can use the fact that f(-1) = 85. So, let's plug in -1 for x:

85 = a((-1)^2 - 4)((-1)^2 + 16)

85 = a(1 - 4)(1 + 16)

85 = a(-3)(17)

85 = a(-51)

Solving for a, we find:

a = -85/51 = -5/3

Putting it all together, we have:

f(x) = (-5/3)(x^2 - 4)(x^2 + 16)

And there you have it! Our nth-degree polynomial function with real coefficients that satisfies the given conditions. I hope I was able to bring a smile to your face while explaining it!

To find an nth-degree polynomial function with real coefficients that satisfies the given conditions, we know that the complex zeros will come in conjugate pairs.

Given that 2i and 4i are zeros, their conjugates are -2i and -4i, respectively.

So, the factors of the polynomial are (x-2i), (x+2i), (x-4i), and (x+4i).

Since these factors are complex conjugates, we can combine them as follows: (x-2i)(x+2i)(x-4i)(x+4i).

We also know that f(-1) = 85.

To find the expression for f(x), we multiply the factors and include the constant term:

f(x) = (x-2i)(x+2i)(x-4i)(x+4i) * k

Expanding this expression, we get:

f(x) = (x^2 + 4)(x^2 + 16) * k

To determine the value of k, we can substitute the x value of -1 into the expression and equate it to 85:

85 = (1^2 + 4)(1^2 + 16) * k

Simplifying the equation, we have:

85 = (5)(17) * k

Now, solve for k:

85 = 85k

k = 1

Therefore, the expression for f(x) is:

f(x) = (x^2 + 4)(x^2 + 16)

Hope this helps! Let me know if you need further assistance.

To find the nth-degree polynomial function satisfying the given conditions, we need to determine the factors of the polynomial.

Since the zeros of the polynomial are 2i and 4i, this means that the factors of the polynomial are (x - 2i) and (x - 4i).

However, since we want a polynomial with real coefficients, the complex conjugate of these factors must also be included. The complex conjugates of 2i and 4i are -2i and -4i, respectively. Therefore, the additional factors are (x + 2i) and (x + 4i).

Now we can find the polynomial function by multiplying these factors together:
f(x) = (x - 2i)(x + 2i)(x - 4i)(x + 4i)

To simplify this expression, we can use the difference of squares pattern:
(a - b)(a + b) = a^2 - b^2

f(x) = [(x - 2i)(x + 2i)][(x - 4i)(x + 4i)]
= [(x^2 - (2i)^2)][(x^2 - (4i)^2)]
= [(x^2 + 4)][(x^2 + 16)]

Now we multiply the remaining binomials:
= (x^2 + 4)(x^2 + 16)
= x^4 + 16x^2 + 4x^2 + 64
= x^4 + 20x^2 + 64

Therefore, the nth-degree polynomial function with real coefficients satisfying the given conditions when n = 4 is:
f(x) = x^4 + 20x^2 + 64.

To check if this is correct, plug in the given value of f(-1) = 85 into the equation:
f(-1) = (-1)^4 + 20(-1)^2 + 64
= 1 + 20 + 64
= 85.

Since f(-1) matches the given value, we can conclude that the polynomial function f(x) = x^4 + 20x^2 + 64 satisfies the conditions.