If $$ f(x)=x^{3}-5 x^{2}-31 x-45 $$ and $$ x-9 $$ is a factor of $$ f(x) $$, then find all of the zeros of $$ f(x) $$ algebraically.

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To find all the zeros of the polynomial $$ f(x) = x^{3} - 5x^{2} - 31x - 45 $$ given that $$ x - 9 $$ is a factor, we can use polynomial long division to divide $$ f(x) $$ by $$ x - 9 $$.

After finding the quotient, we will set it to zero to find the remaining zeros.

Let's perform the polynomial long division.

1. Divide the leading term of $$ f(x) $$ by the leading term of $$ x - 9 $$.
2. Multiply $$ x - 9 $$ by the result from step 1 and subtract from $$ f(x) $$.
3. Repeat the process until we reach a constant or a remainder of zero.

Let's start with the division.

The polynomial long division will be performed as follows:

1. Divide $$ x^3 $$ by $$ x $$ to get $$ x^2 $$.
2. Multiply $$ x^2 $$ by $$ x - 9 $$ to get $$ x^3 - 9x^2 $$.
3. Subtract $$ (x^3 - 9x^2) $$ from $$ f(x) $$.

Now, let's calculate this step by step.

First, we will perform the polynomial long division.
Simplify the expression by following steps:
- step0: Solution:
  $$\frac{\left(x^{3}-5x^{2}-31x-45\right)}{\left(x-9\right)}$$
- step1: Remove the parentheses:
  $$\frac{x^{3}-5x^{2}-31x-45}{x-9}$$
- step2: Factor the expression:
  $$\frac{\left(x-9\right)\left(x^{2}+4x+5\right)}{x-9}$$
- step3: Reduce the fraction:
  $$x^{2}+4x+5$$
After performing the polynomial long division, we find that:

$$
f(x) = (x - 9)(x^{2} + 4x + 5)
$$

Next, we need to find the zeros of the quadratic $$ x^{2} + 4x + 5 $$. We can use the quadratic formula:

$$
x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}
$$

where $$ a = 1 $$, $$ b = 4 $$, and $$ c = 5 $$.

Let's calculate the discriminant $$ b^{2} - 4ac $$ first, and then use the quadratic formula to find the zeros.
Calculate the value by following steps:
- step0: Calculate:
  $$\left(-4\right)-\sqrt{4^{2}-4\times 1\times 5}$$
- step1: Remove the parentheses:
  $$-4-\sqrt{4^{2}-4\times 1\times 5}$$
- step2: Multiply the terms:
  $$-4-\sqrt{4^{2}-20}$$
- step3: Subtract the numbers:
  $$-4-\sqrt{-4}$$
- step4: Simplify the root:
  $$-4-2i$$
Calculate or simplify the expression $$ (-4) + \sqrt(4^2 - 4*1*5) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\left(-4\right)+\sqrt{4^{2}-4\times 1\times 5}$$
- step1: Remove the parentheses:
  $$-4+\sqrt{4^{2}-4\times 1\times 5}$$
- step2: Multiply the terms:
  $$-4+\sqrt{4^{2}-20}$$
- step3: Subtract the numbers:
  $$-4+\sqrt{-4}$$
- step4: Simplify the root:
  $$-4+2i$$
Calculate or simplify the expression $$ 4^2 - 4*1*5 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$4^{2}-4\times 1\times 5$$
- step1: Multiply the terms:
  $$4^{2}-20$$
- step2: Evaluate the power:
  $$16-20$$
- step3: Subtract the numbers:
  $$-4$$
The discriminant $$ b^{2} - 4ac $$ is calculated as:

$$
4^2 - 4 \cdot 1 \cdot 5 = -4
$$

Since the discriminant is negative, the quadratic $$ x^{2} + 4x + 5 $$ has complex roots.

Using the quadratic formula, we find the roots:

$$
x = \frac{-4 \pm \sqrt{-4}}{2 \cdot 1} = \frac{-4 \pm 2i}{2} = -2 \pm i
$$

Thus, the zeros of $$ f(x) $$ are:

1. $$ x = 9 $$ (from the factor $$ x - 9 $$)
2. $$ x = -2 + i $$
3. $$ x = -2 - i $$

In summary, the zeros of $$ f(x) $$ are:

$$
x = 9, \quad x = -2 + i, \quad x = -2 - i
$$
The zeros of $$ f(x) $$ are $$ x = 9 $$, $$ x = -2 + i $$, and $$ x = -2 - i $$.

If and is a factor of , then find all of the zeros of
algebraically.

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