Q. \(x^2 – 2x – 8\)

Step 1: Start with the expression

We are given the expression:

\[
x^2 – 2x – 8
\]

Step 2: Factor the quadratic

To factor a quadratic of the form \(x^2 + bx + c\), we look for two numbers whose product is \(c\) and whose sum is \(b\).

Here, \(b = -2\) and \(c = -8\).

Step 3: Find numbers that multiply to \(-8\) and add to \(-2\)

We list pairs of integers that multiply to \(-8\):

\(1 \cdot (-8) = -8\)

\(2 \cdot (-4) = -8\)

\((-1) \cdot 8 = -8\)

\((-2) \cdot 4 = -8\)

Now check which pair adds to \(-2\):

\(2 + (-4) = -2\)

So the numbers we need are \(2\) and \(-4\).

Step 4: Write the factored form

Using those numbers, we split the middle term \(-2x\) into \(2x\) and \(-4x\):

\[
x^2 – 2x – 8 = x^2 + 2x – 4x – 8
\]

Step 5: Factor by grouping

Group the terms into two pairs:

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

Factor each group:

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

Now combine the results:

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

Factor out the common factor \((x + 2)\):

\[
x(x + 2) – 4(x + 2) = (x + 2)(x – 4)
\]

Final Answer

The expression \(x^2 – 2x – 8\) factors as:

\[
x^2 – 2x – 8 = (x + 2)(x – 4)
\]