If x+6 is a factor x^2-kx-48, what is the value of k

1) (x+6)(Y) = x^2-kx-48

2) Y equals to x-8 as (+6) times (-8) equals to -48

3) Substituting Y to x-8,
(x+6)(x-8) = x^2-2x-48

4) Comparing the Right Hand equation from (3) to x^2-kx-48, we know that k=2
Note: be careful not to say that k=(-2) because in the equation itself it's -kx, so (-kx)=(-2x) yields k=2

ok I'm not sure if this is the formal way to think about it but this is how I've always done it:

(x+6)(x+?) = x^2 - kx - 48

ok so the 2 normal numbers 6 & ? have to multiply to be -48.
therefore what is ?
6 * ? = -48
? = -8

so the equation is:
(x+6)(x-8)
now merely expand
x^2 -2x -48

therefore -2x = - kx
divide both sides by -x
2 = k

It would be k=2.

(x+6)(x-8) when multiplied through is x^2 - 2x - 48.

K = 2