Find the real and complex zeros of the following function. Please show all of your work.

f(x) = x^3 + 2x^2 - 6x + 8

f(±1) ≠ 0

f(±2) ≠ 0
but f(-4) = 0
so x+4 is a factor

I then used synthetic division to get
(x+4)(x^2 - 2x + 2) = 0

the real root is x = -4

Use the quadratic formula to get the two complex roots of
1+i and 1-i

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

(2x^2+8x)+(2x+8)=x^2(x+4)-2x(x+4)+2(x+4)=
(x+4)(x^2-2x+2)
x+4=0 => x=-4
or
x^2-2x+2=0 => x=1+i or x=1-i

To find the zeros of the function f(x) = x^3 + 2x^2 - 6x + 8, we need to find the values of x for which f(x) equals zero.

Step 1: Check for rational zeros
To check for rational zeros, we use the Rational Root Theorem. According to the theorem, the possible rational zeros of the function f(x) are all the factors of the constant term (8) divided by all the factors of the leading coefficient (1).

The factors of 8 are ±1, ±2, ±4, ±8.
The factors of 1 (leading coefficient) are ±1.

Therefore, the possible rational zeros are ±1, ±2, ±4, ±8.

Step 2: Test the possible rational zeros
To test the possible rational zeros, we can use synthetic division. Starting with the first possible rational zero, we will proceed until we find a zero or exhaust all the possibilities.

Let's start by testing x = 1:

| 1 2 -6 8 |
|_______|
| |

Bring down the coefficient of x^3, which is 1:
| 1 2 -6 8 |
|_____|
| 1

Multiply the previous result by the value of x being tested (1) and add it to the next coefficient:
| 1 2 -6 8 |
|____|
| 1 3

Repeat the process:
| 1 2 -6 8 |
|____|
| 1 3 -3

Finally, we get a remainder of 5. Since the remainder is not zero, x = 1 is not a zero of the function.

Let's test the next possible rational zero, x = -1:
| 1 2 -6 8 |
|_______|
| |

Using synthetic division:
| 1 2 -6 8 |
|_____|
| -1

| 1 2 -6 8 |
|____|
| 1 1

| 1 2 -6 8 |
|____|
| 1 1 -5

Again, we get a remainder of 3. Therefore, x = -1 is not a zero of the function.

We can repeat this process for all the other possible rational zeros (+2, -2, +4, -4, +8, -8).

After testing all the possible rational zeros, we found that none of them are zeros of the function f(x) = x^3 + 2x^2 - 6x + 8.

Therefore, there are no rational zeros. However, there may be complex zeros.

Step 3: Use the complex conjugate theorem
Since all the coefficients of the polynomial are real, any complex zeros must come in conjugate pairs.

Step 4: Apply factoring or the quadratic formula
To find the complex zeros, we can either factor the polynomial or use the quadratic formula if we have a quadratic factor.

However, after testing all the possible rational zeros, we found that none of them are zeros. Therefore, the polynomial cannot be factored into linear factors.

Next, let's check if there is any quadratic factor.

We can use long division or synthetic division to divide the original polynomial by (x - a) where a is a possible zero.

Using synthetic division, let's try (x - 2):

| 1 2 -6 8 |
|_______|
| |

| 1 2 -6 8 |
|____|
| 2

| 1 2 -6 8 |
|____|
| 2 4

| 1 2 -6 8 |
|____|
| 2 4 -2

The remainder is not zero, so (x - 2) is not a factor of the polynomial.

Hence, we can conclude that there are no quadratic factors and no rational nor complex zeros for the function f(x) = x^3 + 2x^2 - 6x + 8.

Therefore, the function f(x) = x^3 + 2x^2 - 6x + 8 has no real or complex zeros.

To find the real and complex zeros of the function f(x) = x^3 + 2x^2 - 6x + 8, we can use the method of factoring and the quadratic formula. Here are the steps to follow:

Step 1: Factor out the common factor (if any)
Check if there is any common factor for all the terms in the equation. In this case, we can see that there is no common factor that can be factored out.

Step 2: Try to find any rational zeros
The rational zeros are the possible values of x that would make the polynomial equation equal to zero. We can use the Rational Root Theorem to find the rational zeros. According to the theorem, any rational zero of a polynomial equation in the form f(x) = a_nx^n + a_(n-1)x^(n-1) + ... + a_0 will be of the form p/q, where p is a factor of the constant term a_0 and q is a factor of the leading coefficient a_n.

In our equation, the leading coefficient is 1, and the constant term is 8. The factors of 8 are ±1, ±2, ±4, and ±8, and the factors of 1 (leading coefficient) are ±1. So, the possible rational zeros are ±1, ±2, ±4, and ±8.

Step 3: Test the possible rational zeros
We can test each possible rational zero by substituting it into the equation f(x) and check if it equals zero. Let's start with x = 1.

f(1) = (1)^3 + 2(1)^2 - 6(1) + 8
= 1 + 2 - 6 + 8
= 5

The result is not zero, so x = 1 is not a zero of f(x). We can proceed to test the other possible rational zeros in the same way.

f(-1) = (-1)^3 + 2(-1)^2 - 6(-1) + 8
= -1 + 2 + 6 + 8
= 15

f(2) = (2)^3 + 2(2)^2 - 6(2) + 8
= 8 + 8 - 12 + 8
= 12

f(-2) = (-2)^3 + 2(-2)^2 - 6(-2) + 8
= -6 + 8 + 12 + 8
= 22

f(4) = (4)^3 + 2(4)^2 - 6(4) + 8
= 64 + 32 - 24 + 8
= 80

f(-4) = (-4)^3 + 2(-4)^2 - 6(-4) + 8
= -64 + 32 + 24 + 8
= 0

We have found a zero! f(-4) = 0, which means x = -4 is a zero of f(x).

Step 4: Use long division to find the quadratic equation
Now that we have found one zero, x = -4, we can use long division to divide the original equation f(x) by (x + 4). This will give us a quadratic equation which is easier to solve.

x^2 - 2x - 4
-----------------------------
x + 4 | x^3 + 2x^2 - 6x + 8
-(x^3 + 4x^2)
---------------
-2x^2 - 6x + 8
-(-2x^2 - 8x)
---------------
2x + 8
-(2x + 8)
---------------
0

The result of the division is x^2 - 2x - 4.

Step 5: Solve the quadratic equation
Now, we have the quadratic equation x^2 - 2x - 4. We can solve this equation by factoring or using the quadratic formula. Let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

For our equation, a = 1, b = -2, and c = -4.

x = (-(-2) ± √((-2)^2 - 4(1)(-4))) / (2(1))
= (2 ± √(4 + 16)) / 2
= (2 ± √(20)) / 2
= (2 ± 2√5) / 2
= 1 ± √5

So, the complex zeros of the function are x = 1 + √5 and x = 1 - √5.

In summary, the real zero is x = -4, and the complex zeros are x = 1 + √5 and x = 1 - √5.