Form a polynomial f(x) with real coefficients having the given degree and zeros.

Degree 5;
Zeros: -3; -i; -6+i
F(x)=a ( )

Degree 5, huh? That's a big one! Alright, let's put our funny math hats on and create a polynomial that fits the bill.

Given the zeros, we have:
x = -3,
x = -i (imaginary zero),
x = -6 + i (complex zero).

Now, remember that complex zeros come in conjugate pairs. So, if we have -i as a zero, we also have its conjugate, which is i. Similarly, if we have -6 + i as a zero, we also have its conjugate, which is -6 - i.

So, putting it all together, our polynomial is:

f(x) = a(x - (-3))(x - (-i))(x - i)(x - (-6 + i))(x - (-6 - i))

Simplifying this further, we get:

f(x) = a(x + 3)(x + i)(x - i)(x + 6 - i)(x + 6 + i)

And there you have it! A polynomial of degree 5 with the given zeros. Don't worry, it's not as complicated as it looks – it's just math trying to put on its clown suit!

To form a polynomial with degree 5 and the given zeros, we can use the fact that for a polynomial with real coefficients, if a complex number is a zero, then its conjugate is also a zero.

Given zeros: -3, -i, -6+i

Since -3 is a zero, we can form a factor (x + 3).
Since -i and its conjugate i are zeros, we can form a factor (x + i)(x - i) = (x^2 + 1).

Similarly, the complex number -6+i has a conjugate -6-i. So, we can form a factor (x - (-6+i))(x - (-6-i)).

Expanding the last factor, we get: (x + 6 - i)(x + 6 + i) = (x^2 + 12x + 37).

Combining all the factors, the polynomial f(x) is:

f(x) = a(x + 3)(x^2 + 1)(x^2 + 12x + 37).

To form a polynomial with the given degree and zeros, we need to consider that the zeros can be either real or complex conjugate pairs.

Given zeros:
1. -3 (real zero)
2. -i (complex zero)
3. -6 + i (complex zero)

Degree 5 means the polynomial will have five terms. To get the polynomial function, we can use the fact that complex roots occur in conjugate pairs. Therefore, if -i is a zero, then its conjugate, i, must also be a zero. Similarly, if -6 + i is a zero, then its conjugate, -6 - i, must also be a zero.

So, the zeros of the polynomial are: -3, -i, i, -6 + i, and -6 - i.

To find the polynomial, we can multiply the factors corresponding to each zero. The general form of a polynomial with real coefficients is:

f(x) = a(x - r1)(x - r2)(x - r3)(x - r4)(x - r5)

Where a is a constant and r1, r2, r3, r4, and r5 are the zeros.

Now let's substitute our zeros in the above equation and simplify:

f(x) = a(x + 3)(x + i)(x - i)(x - (-6 + i))(x - (-6 - i))

Expanding and simplifying each term:

f(x) = a(x + 3)(x^2 + 1)(x + 6 - i)(x + 6 + i)

f(x) = a(x + 3)(x^2 + 1)(x^2 + 12x + 37)

Now, the polynomial f(x) with degree 5 and the given zeros is:

f(x) = a(x + 3)(x^2 + 1)(x^2 + 12x + 37)

Complex numbers always appear as conjugate pairs, so if you have -i, then you also have +i

and if you have -6+i, there will also be -6 - i

so we know we have factors of (x+3) , (x^2 + 1) and two more

I will use the sum and product rule to find the other
sum of -6+i and -6 - i = -12
product of the above is 36 - i^2 = 36 + 1 = 37
resulting in the quadratic factor
x^2 + 12x + 37
You could also expand (x -(-6+i))(x - (-6-i)) and get the same result

so f(x) = (x+3)(x^2 + 1)(x^2 + 12x + 37)

notice, if expanded this will give you a 5th degree polynomial. If you have to expand it, do it very carefully and patiently.