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.**

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**

To find k if x^2+ kx - 6 = (x - 2)(x + 3).

Solution:

We expand the right side and solve the equation.

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

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

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

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

Subtract x^2-6 from both sides:

kx = x.

**So k = 1.**