write as the product of linear factors: x^2 + 6x - 2

To express the quadratic \( x^2 + 6x - 2 \) as the product of linear factors, we first need to find its roots using the quadratic formula:

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

For the equation \( ax^2 + bx + c \), we have:

- \( a = 1 \)
- \( b = 6 \)
- \( c = -2 \)

Substituting these values into the formula:

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

Calculating the discriminant:

\[
b^2 - 4ac = 36 + 8 = 44
\]

Now, substituting back:

\[
x = \frac{{-6 \pm \sqrt{44}}}{2}
\]

Since \( \sqrt{44} = 2\sqrt{11} \), we can simplify further:

\[
x = \frac{{-6 \pm 2\sqrt{11}}}{2} = -3 \pm \sqrt{11}
\]

Thus, the roots are:

\[
x_1 = -3 + \sqrt{11}, \quad x_2 = -3 - \sqrt{11}
\]

Now, we can write the quadratic as the product of its linear factors:

\[
(x - x_1)(x - x_2) = \left(x - (-3 + \sqrt{11})\right)\left(x - (-3 - \sqrt{11})\right)
\]

This simplifies to:

\[
(x + 3 - \sqrt{11})(x + 3 + \sqrt{11})
\]

So, the final expression for the quadratic in terms of linear factors is:

\[
(x + 3 - \sqrt{11})(x + 3 + \sqrt{11})
\]

If you have any questions or want further details, feel free to ask!

### Related Questions:
1. How can we check the correctness of these factors?
2. What is the graphical representation of this quadratic?
3. Can you solve \( x^2 + 6x - 2 = 0 \) using another method?
4. How do the roots relate to the vertex of the parabola?
5. What is the significance of the discriminant in this context?

**Tip:** Always check your roots by substituting them back into the original equation to ensure they satisfy it.

Explore the step-by-step solution to factor the quadratic equation x^2 + 6x - 2 into linear factors. This guide covers the quadratic formula, root calculation, and algebraic simplification, making it ideal for high school students in Grades 9-11.

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