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The problem gives the polynomial:

\[
h(x) = x^3 - 4x^2 - 8x + 8
\]

It also mentions that \( -2 \) is a zero of the polynomial, meaning \( x = -2 \) is a root. We are asked to express \( h(x) \) as a product of linear factors.

### Step 1: Use Synthetic Division
Since \( -2 \) is a zero, we can divide the polynomial \( h(x) \) by \( x + 2 \) using synthetic division.

Let's perform synthetic division on \( x^3 - 4x^2 - 8x + 8 \) by \( x + 2 \):

#### Coefficients:
The coefficients of the polynomial are: 
\[
1, -4, -8, 8
\]

#### Synthetic Division Process:

\[
\begin{array}{r|rrrr}
-2 & 1 & -4 & -8 & 8 \\
   &   & -2 & 12 & -8 \\
\hline
   & 1 & -6 & 4  & 0 \\
\end{array}
\]

The result of the division is \( x^2 - 6x + 4 \), with no remainder, as expected since \( -2 \) is a root.

### Step 2: Factor the Quadratic

Now we need to factor the quadratic \( x^2 - 6x + 4 \). We can use the quadratic formula:

\[
x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(4)}}{2(1)}
\]

\[
x = \frac{6 \pm \sqrt{36 - 16}}{2}
\]

\[
x = \frac{6 \pm \sqrt{20}}{2}
\]

\[
x = \frac{6 \pm 2\sqrt{5}}{2}
\]

\[
x = 3 \pm \sqrt{5}
\]

Thus, the quadratic factors as:

\[
(x - (3 + \sqrt{5}))(x - (3 - \sqrt{5}))
\]

### Step 3: Write the Polynomial as a Product of Linear Factors

Since \( -2 \) is a zero, the factor corresponding to it is \( (x + 2) \). Therefore, the complete factorization of \( h(x) \) is:

\[
h(x) = (x + 2)(x - (3 + \sqrt{5}))(x - (3 - \sqrt{5}))
\]

This is the polynomial \( h(x) \) expressed as a product of linear factors.

Would you like more details or have any questions?

Here are 5 related questions you can explore:
1. How does synthetic division differ from polynomial long division?
2. What are the advantages of expressing polynomials as products of linear factors?
3. Can all cubic polynomials be factored into linear factors over real numbers?
4. How does the quadratic formula help in factoring second-degree polynomials?
5. What are the implications of irrational roots in real-world applications?

**Tip:** The quadratic formula is always useful for factoring quadratics when other methods like factoring by inspection or completing the square don't work easily!

For the polynomial h(x) = x^3 - 4x^2 - 8x + 8, express it as a product of linear factors, given that -2 is a zero.

Learn how to express the polynomial h(x) = x^3 - 4x^2 - 8x + 8 as a product of linear factors. Follow the steps using synthetic division and the quadratic formula. This problem is suitable for students in Grades 10-12.

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