To factor the quadratic expression $9x^2 - 13x - 4$ by grouping, let's follow these steps:

Step 1: Multiply the coefficient of $x^2$ by the constant term

We multiply the coefficient of $x^2$ (which is 9) by the constant term (which is -4):

$9 \times -4 = -36$

Step 2: Find two numbers that multiply to -36 and add to -13

We need two numbers whose product is $-36$ and whose sum is $-13$. These numbers are $-18$ and $2$, because:

$-18 \times 2 = -36 \quad \text{and} \quad -18 + 2 = -13$

Step 3: Rewrite the middle term using the two numbers

Now, we can rewrite the quadratic expression $9x^2 - 13x - 4$ as:

$9x^2 - 18x + 2x - 4$

Step 4: Group the terms and factor each group

Group the terms into two pairs:

$(9x^2 - 18x) + (2x - 4)$

Factor out the greatest common factor (GCF) from each pair:

$9x(x - 2) + 2(x - 2)$

Step 5: Factor out the common binomial factor

Now, we can factor out the common binomial factor $(x - 2)$:

$(9x + 2)(x - 2)$

Thus, the factored form of $9x^2 - 13x - 4$ is:

$(9x + 2)(x - 2)$

Would you like a more detailed explanation or have any questions about this process?

Here are some related questions you might find interesting:

1. How do you factor quadratics when the leading coefficient is 1?
2. What if the quadratic has a perfect square trinomial structure?
3. Can this method be applied to polynomials with more than two terms?
4. How can you verify the factorization is correct using the distributive property?
5. What happens if the two numbers we find do not exist for a given quadratic?

Tip: Always check for a greatest common factor (GCF) before starting to factor a quadratic, as this can simplify the problem significantly.