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

4x^2 - 19x - 5

=  4x^2 - 20x + x - 5

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

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

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

(x-5)(4x+1) = 4x² – 19x – 5

B-75

Step-by-step explanation:

110-35+75

To solve this problem you will have subtract B from A.