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We are tasked with factoring the quadratic expression \( 4x^2 - 19x - 5 \). 

To factor this expression, we look for two numbers that:
1. Multiply to give the product of the coefficient of \( x^2 \) (which is 4) and the constant term (which is -5). That is, we need two numbers whose product is \( 4 \times (-5) = -20 \).
2. Add up to give the middle coefficient, which is -19.

Let's find these numbers.

The factors of -20 that add up to -19 are -20 and 1, because:
\[
-20 \times 1 = -20
\quad \text{and} \quad
-20 + 1 = -19
\]

Now, we rewrite the middle term \( -19x \) using these two numbers:
\[
4x^2 - 20x + x - 5
\]

Next, we group the terms and factor by grouping:
\[
(4x^2 - 20x) + (x - 5)
\]
\[
4x(x - 5) + 1(x - 5)
\]

Now, we factor out the common factor \( (x - 5) \):
\[
(4x + 1)(x - 5)
\]

Thus, the factored form of \( 4x^2 - 19x - 5 \) is:
\[
(4x + 1)(x - 5)
\]

Let me know if you'd like more details or have any questions!

Here are five related questions:
1. How do you factor trinomials where the leading coefficient is not 1?
2. Can this factoring technique be applied to all quadratic expressions?
3. How can you check if the factorization is correct?
4. What are the roots of the equation \( 4x^2 - 19x - 5 = 0 \)?
5. How would the factorization change if the signs in the quadratic were different?

**Tip:** Always double-check your factorization by expanding the factors to ensure they match the original expression.

Learn how to factor the quadratic expression 4x^2 - 19x - 5 step-by-step. This involves using polynomial factorization techniques, including factoring by grouping, to break down the quadratic into two binomials. Suitable for students in Grades 9-10.

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