Given that 9x² - 30x - 40 + k is a perfect square, find the value of k.

A perfect square trinomial is of the form $(ax \pm b)^2 = a^2x^2 \pm 2abx + b^2$.
Comparing $9x^2 - 30x - 40 + k$ with this form, we can identify $a^2x^2 = 9x^2$ and $2abx = -30x$.

From $a^2x^2 = 9x^2$, we get $a = 3$.
From $2abx = -30x$, substitute $a=3$:
$2(3)bx = -30x$
$6bx = -30x$
$6b = -30$
$b = -5$ (or we can consider $b=5$ and the form $(ax-b)^2$)

If $a=3$ and $b=5$, the perfect square is $(3x - 5)^2$.
Expanding $(3x - 5)^2$:

The given expression is $9x^2 - 30x - 40 + k$.
For this to be a perfect square, its constant term must be 25.
So, we set $-40 + k = 25$.