We are given that x^2+ kx - 6 = (x - 2)(x + 3)

Now x^2+ kx - 6 = (x - 2)(x + 3)

=> x^2 + kx - 6 = x^2 -2x + 3x - 6

=> x^2 + kx - 6 = x^2 + x - 6

canceling the common terms

=> kx = x

=> k = 1

Therefore k = 1

Given that x^2 + kx - 6 = (x-2)(x+3)

We need to find k.

First we will need to open the brackets on the left side and then compare the terms.

==> x^2 + kx -6 = x^2 -2x + 3x - 6

==> x^2 + kx -6 = x^2 + x -6

Now we will add 6 to both sides.

==> x^2 + kx = x^2 + x

Now we will subtract x^2

==> kx = x

Now we will subtract x from both sides.

==> kx -x = 0

Now we will factor x.

==> x (k-1) = 0

x can not be zero.

Then k-1 = 0  ==> k= 1

Then the value of k is 1.