To find the value of [tex]\( k \)[/tex] such that the equation [tex]\((x - k)(x + 4k) = x^2 + 6x - 16\)[/tex] holds true, we will follow these steps:
1. Expand the left-hand side of the equation:
[tex]\[
(x - k)(x + 4k)
\][/tex]
Using the distributive property (FOIL method) to expand:
[tex]\[
(x - k)(x + 4k) = x(x + 4k) - k(x + 4k)
\][/tex]
[tex]\[
= x^2 + 4kx - kx - 4k^2
\][/tex]
Simplifying, we get:
[tex]\[
= x^2 + (4k - k)x - 4k^2
\][/tex]
[tex]\[
= x^2 + 3kx - 4k^2
\][/tex]
2. Set the expanded left-hand side equal to the right-hand side:
[tex]\[
x^2 + 3kx - 4k^2 = x^2 + 6x - 16
\][/tex]
3. Eliminate common terms:
Subtract [tex]\( x^2 \)[/tex] from both sides:
[tex]\[
3kx - 4k^2 = 6x - 16
\][/tex]
4. Compare coefficients:
The equation should hold for all values of [tex]\( x \)[/tex], so we compare the coefficients of [tex]\( x \)[/tex] and the constant term on both sides. This gives us a system of equations:
From the coefficient of [tex]\( x \)[/tex]:
[tex]\[
3k = 6
\][/tex]
Solving this, we get:
[tex]\[
k = \frac{6}{3} = 2
\][/tex]
From the constant term:
[tex]\[
-4k^2 = -16
\][/tex]
[tex]\[
4k^2 = 16
\][/tex]
Solving this, we get:
[tex]\[
k^2 = \frac{16}{4} = 4
\][/tex]
Taking the square root of both sides, we get:
[tex]\[
k = \pm 2
\][/tex]
So, the valid values for [tex]\( k \)[/tex] based on the system we have are [tex]\( k = 2 \)[/tex] or [tex]\( k = -2 \)[/tex].
After carefully comparing the coefficients and constants again, and considering the relationships established initially, we also notice:
[tex]\[
3k = 6 \Rightarrow k = 2
\][/tex]
So, an alternative value for k, when expressed solely in terms of x is:
[tex]\[ k = \frac{3x}{4} - 2 \][/tex]
Thus, taking into account all possibilities, we have:
[tex]\[ k = 2 \][/tex]
or
[tex]\[ k = \frac{3x}{4} - 2 \][/tex]
Therefore, we box our final solutions:
[tex]\[
k = \boxed{2}
\][/tex]
and
[tex]\[
k = \boxed{\frac{3x}{4} - 2}
\][/tex]
