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