## Answer :

1.

**Identify the quadratic equation:**The given polynomial is [tex]\( x^2 + 2x - 8 \)[/tex].

2.

**Recognize the standard form of a quadratic equation:**A quadratic equation is typically written in the form [tex]\( ax^2 + bx + c = 0 \)[/tex]. For the polynomial [tex]\( x^2 + 2x - 8 \)[/tex]:

- [tex]\( a = 1 \)[/tex]

- [tex]\( b = 2 \)[/tex]

- [tex]\( c = -8 \)[/tex]

3.

**Use the quadratic formula to find the roots:**The quadratic formula is given by:

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

Let's substitute the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the quadratic formula.

4.

**Calculate the discriminant:**The discriminant [tex]\( \Delta \)[/tex] is the part under the square root in the quadratic formula:

[tex]\[ \Delta = b^2 - 4ac \][/tex]

Plugging in the coefficients:

[tex]\[ \Delta = 2^2 - 4(1)(-8) = 4 + 32 = 36 \][/tex]

5.

**Find the roots using the quadratic formula:**

[tex]\[ x = \frac{-2 \pm \sqrt{36}}{2 \cdot 1} \][/tex]

Simplifying inside the square root:

[tex]\[ \sqrt{36} = 6 \][/tex]

So, we have two possible roots:

[tex]\[ x_1 = \frac{-2 + 6}{2} = \frac{4}{2} = 2 \][/tex]

[tex]\[ x_2 = \frac{-2 - 6}{2} = \frac{-8}{2} = -4 \][/tex]

6.

**Write the factorization of the quadratic polynomial:**Once we have the roots [tex]\( x_1 = 2 \)[/tex] and [tex]\( x_2 = -4 \)[/tex], the factorization of the polynomial [tex]\( x^2 + 2x - 8 \)[/tex] can be written as:

[tex]\[ (x - x_1)(x - x_2) = (x - 2)(x + 4) \][/tex]

7.

**Match the factorization with the given options:**

- A. [tex]\( (x+4)(x-2) \)[/tex]

- B. [tex]\( (x-8)(x+1) \)[/tex]

- C. [tex]\( (x+6)(x-4) \)[/tex]

- D. [tex]\( (x-4)(x+2) \)[/tex]

The correct factorization of [tex]\( x^2 + 2x - 8 \)[/tex] is [tex]\( (x + 4)(x - 2) \)[/tex], which corresponds to option A.

Therefore, the factorization of [tex]\( x^2 + 2x - 8 \)[/tex] is:

[tex]\[ \boxed{(x + 4)(x - 2)} \][/tex]