Which factorization of 4x2 - 19x - 5 is correct?

1st find the roots (or the zero) of the quadratic equation:

4x² - 19x - 5

[-b + √ (b²-4. a. c) ]/2

x₁ = [-b + √ (b²-4. a. c) ]/2 and x₂ = [-b - √ (b²-4. a. c) ]/2

x₁ = 5 and x₂ = - 1/4

ax² + bx + c = a (x-x₁) (x-x₂)

Then a (x-x₁) (x-x₂) = 4 (x-5) (x+1/4) (put second term at same denominator:

Or 4 (x-5) (4x+1) / 4. Now simplify by 4, final answer:

(x-5) (4x+1) = 4x² - 19x - 5

4x^2 - 19x - 5

= 4x^2 - 20x + x - 5

= 4x (x - 5) + 1 (x - 5)

= (4x + 1) (x - 5)