回答所有问题。

1. Differentiate $f(x)$ using the product rule.

$f(x) = (2x - x^2)e^x$
Let $u = 2x - x^2$ and $v = e^x$.
$\frac{du}{dx} = 2 - 2x$
$\frac{dv}{dx} = e^x$
$f'(x) = u\frac{dv}{dx} + v\frac{du}{dx}$
$f'(x) = (2x - x^2)e^x + e^x(2 - 2x)$

2. Simplify the derivative.

$f'(x) = e^x(2x - x^2 + 2 - 2x)$
$f'(x) = e^x(2 - x^2)$

3. Set up and solve the inequality for a decreasing function.

For $f(x)$ to be a decreasing function, $f'(x) < 0$.
$e^x(2 - x^2) < 0$
Since $e^x > 0$ for all real $x$, we must have:
$2 - x^2 < 0$
$x^2 > 2$
$x < -\sqrt{2}$ or $x > \sqrt{2}$

1. Find the gradient of the line parallel to the tangent

The line is $y + 2x - 5 = 0$, which can be rewritten as $y = -2x + 5$.
The gradient of this line is $-2$.

2. Equate the gradient function at $x = 2$ to the gradient of the line

Since the tangent at $(2, -2)$ is parallel to the line, its gradient is also $-2$.
Substitute $x = 2$ into the gradient function:
$2(p + 1)(2) + 2 = -2$

3. Solve for $p$

$4(p + 1) = -4$
$p + 1 = -1$
$p = -2$