Q. \(x^2-x-20\)

Step 1: Rewrite the expression in a standard polynomial form.

The problem is to work with the quadratic expression \(x^2 – x – 20\).

Step 2: Factor the quadratic.

We want to factor \(x^2 – x – 20\) into the form \((x – a)(x + b)\), where \(a\) and \(b\) are positive numbers. This is because the product \((x – a)(x + b)\) will produce:

\[
(x – a)(x + b) = x^2 + (b-a)x – ab.
\]

We need this to match \(x^2 – x – 20\). So we match coefficients:

\[
b – a = -1,\quad ab = 20.
\]

Step 3: Find two numbers that multiply to \(20\) and differ by \(1\).

The factor pairs of \(20\) are:

\[
1 \cdot 20,\quad 2 \cdot 10,\quad 4 \cdot 5.
\]

Now check which pair gives a difference of \(1\):

\[
2 \text{ and } 1 \text{ do this, since } 2 – 1 = 1.
\]

To make \(b – a = -1\), we need \(a = 2\) and \(b = 1\). Then \(ab = 2 \cdot 1 = 2\), which is not correct for \(20\). So instead we use the factorization with negative numbers correctly.

We look for numbers that multiply to \(20\) and add to \(-1\) (since the middle term is \(-x\)). That means we need two numbers \(m\) and \(n\) such that:

\[
m \cdot n = -20,\quad m + n = -1.
\]

Choose \(m = 4\) and \(n = -5\). Then:

\[
4 \cdot (-5) = -20,\quad 4 + (-5) = -1.
\]

Step 4: Use those numbers to factor.

Substitute \(4\) and \(-5\) into the factoring pattern:

\[
x^2 – x – 20 = (x + 4)(x – 5).
\]

Final Answer:

\[
x^2 – x – 20 = (x + 4)(x – 5).
\]