x^2-4x-5=(x+ 1)(x-5)

The quadratic term coefficient is 1, the linear term coefficient is-1, and the constant term is -5, then

a= 1 b= 1

×

c= 1 d=-5

The quadratic term a*c= 1* 1= 1, the constant term b*d 1*(-5)=-5, and the linear term a * d+c * b =1* (-5).

x^2-x-5=(ax+b)(cx+d)=(x+ 1)(x-5)

For x 2 -7x-10, cross factorization cannot be used, because the sum of two real factors of-10 cannot be equal to -7.

Only radical methods can be used:

{x-[-b+√(b^2-4ac)]a}{x-[-b-√(b^2-4ac)]a}

x^2-7x- 10=[x-(7+√89)][x-(7-√89)]