A polynomial function f(x) with real coefficients has the given degree, zeros, and solution point.

Degree: 3
Zeros: -2,2+2√2i
Solution Point: f(−1) = −68

(a) Write the function in completely factored form.

(b) Write the function in polynomial form.

Help Please my teacher doesn't really teach!!!

you have one complex root, but they always come as conjugate pairs

The other must be 2-2√2i
You also know that one factor is (x+2)

the sum of the complex roots
= 2+2√2i + 2-2√2i = 4
product of the complex roots
= (2+2√2i)(2-2√2i) = 4 + 8 = 12

So the quadratic factor is x^2 - 4x + 12

f(x) = a(x+2)(x^2 - 4x + 12)
we also have a point (-1,-68)

-68 = a(1)(17)
a = -4

f(x) = -4(x+2)(x^2 - 4x + 12)

I will leave it up to you to expand it.

Check of my answer:
https://www.wolframalpha.com/input/?i=-4(x%2B2)(x%5E2+-+4x+%2B+12)%3D0

write a polynomial function of minimum degree with real coefficients whose zeros include those listed. write the polynomial in standard form calculator

Sure, I'd be happy to help you!

(a) To write the function in completely factored form, we can use the given zeros. The zeros are -2, 2+2√2i, and its complex conjugate -2√2i.

So, the completely factored form would be:
f(x) = (x + 2)(x - 2 - 2√2i)(x - 2 + 2√2i)

(b) To write the function in polynomial form, we can multiply out the terms in the factored form:
f(x) = (x + 2)(x^2 - 4x + 8)

So, the polynomial form of the function would be:
f(x) = x^3 - 4x^2 + 8x + 2x^2 - 8x + 16

Simplifying further, we have:
f(x) = x^3 - 2x^2 + 8

I hope this helps! Let me know if you have any more questions.

No worries! I'll guide you through the steps to find the function in both completely factored form and polynomial form.

(a) To write the function in completely factored form, we use the zeros of the function. Given that the zeros are -2, 2+2√2i, and 2-2√2i, we can start by setting up factors for each of these zeros.

The factor for -2 can be written as (x - (-2)), which simplifies to (x + 2).

Now, let's focus on the complex zeros 2+2√2i and 2-2√2i. Complex zeros always come in conjugate pairs, which means if we have one, we'll have the other. The complex conjugate of 2+2√2i is 2-2√2i, and vice versa. So the factors for the complex zeros are (x - (2 + 2√2i)) and (x - (2 - 2√2i)).

Next, we multiply these factors together to get the completely factored form. Here's the expression:

f(x) = (x + 2)(x - (2 + 2√2i))(x - (2 - 2√2i))

(b) To write the function in polynomial form, we need to expand and simplify the expression obtained in part (a) by multiplying all the factors:

f(x) = (x + 2)(x - (2 + 2√2i))(x - (2 - 2√2i))

To simplify this expression, we need to use the conjugate property of complex numbers. When we multiply a complex number by its conjugate, we eliminate the imaginary part.

Let's multiply the conjugates first:

(x - (2 + 2√2i))(x - (2 - 2√2i))

Using the formula (a + b)(a - b) = a^2 - b^2, we can simplify it to:

(x^2 - (2 + 2√2i)^2)

Expanding the squared term:

(x^2 - (2^2 + 2 * 2 * 2√2i + (2√2i)^2))

Simplifying and using the fact that (2√2i)^2 = 4(√2i)^2 = 4 * 2i = 8i:

(x^2 - (4 + 8√2i - 8i))

(x^2 - 4 - 8√2i + 8i)

Now we can multiply all the remaining factors:

f(x) = (x + 2)(x^2 - 4 - 8√2i + 8i)

Expand and simplify further:

f(x) = x^3 + 2x^2 - 4x - 8√2i - 8i - 16 - 32√2i + 32i

Finally, simplify the terms:

f(x) = x^3 + 2x^2 - 4x - 16 + (24 - 8√2i + 32i - 32√2i)

f(x) = x^3 + 2x^2 - 4x - 16 + (-8√2i + 32i - 32√2i + 24)

Combining like terms:

f(x) = x^3 + 2x^2 - 4x - 8√2i - 32√2i + 32i + 24 - 16

f(x) = x^3 + 2x^2 - 4x - (8 + 32)√2i + 32i + 8

f(x) = x^3 + 2x^2 - 4x - 40√2i + 32i + 8

So, the polynomial form of the function is:

f(x) = x^3 + 2x^2 - 4x - 40√2i + 32i + 8

I hope this explanation helps you understand how to find the completely factored form and polynomial form of a given polynomial function with specific zeros and solution points!