For the polynomial below, 3 is a zero. $$ f(x)=x^{3}-7 x^{2}+9 x+9 $$ Express $$ f(x) $$ as a product of linear factors. $$ f(x)=\square $$

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We start with
$$
f(x)=x^3-7x^2+9x+9.
$$
Since $$3$$ is a zero, $$(x-3)$$ is a factor of $$f(x)$$.

**Step 1. Divide $$f(x)$$ by $$(x-3)$$:**

Using synthetic division with $$3$$:

- Write the coefficients: $$1,\, -7,\, 9,\, 9$$.
- Bring down the $$1$$.
- Multiply $$1$$ by $$3$$ and add to $$-7$$ to get: $$-7+3=-4$$.
- Multiply $$-4$$ by $$3$$ and add to $$9$$ to get: $$9-12=-3$$.
- Multiply $$-3$$ by $$3$$ and add to $$9$$ to get: $$9-9=0$$.

The quotient is the quadratic
$$
x^2-4x-3.
$$

**Step 2. Factor the quadratic $$x^2-4x-3$$:**

To factor $$x^2-4x-3$$, we use the quadratic formula:
$$
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.
$$
Here, $$a=1$$, $$b=-4$$, and $$c=-3$$. Thus,
$$
x=\frac{-(-4)\pm\sqrt{(-4)^2-4\cdot1\cdot(-3)}}{2\cdot1}=\frac{4\pm\sqrt{16+12}}{2}=\frac{4\pm\sqrt{28}}{2}.
$$
Since $$\sqrt{28}$$ simplifies to $$2\sqrt{7}$$, we have:
$$
x=\frac{4\pm2\sqrt{7}}{2}=2\pm\sqrt{7}.
$$

Thus, the quadratic factors as:
$$
x^2-4x-3=(x-(2+\sqrt{7}))(x-(2-\sqrt{7})).
$$

**Step 3. Write $$f(x)$$ as a product of linear factors:**

Combining the factor $$(x-3)$$ with the factors obtained from the quadratic, we have:
$$
f(x)=(x-3)(x-(2+\sqrt{7}))(x-(2-\sqrt{7})).
$$

So, the final answer is:
$$
f(x)=(x-3)(x-(2+\sqrt{7}))(x-(2-\sqrt{7})).
$$
$$ f(x) = (x - 3)(x - (2 + \sqrt{7}))(x - (2 - \sqrt{7})) $$

For the polynomial below, 3 is a zero.

Express as a product of linear factors.

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For the polynomial below, 3 is a zero. Express as a product of linear factors.