Integrate $\left(\frac{6}{e^{3-x}}\right)^2$ with respect to $x$. Explain, with working, why the equation $2x^3 + 3x^2 + 2x + 8 = 0$ has only 1 real root. Hence find this root.

1. Simplify the integrand using the laws of indices

$\left(\frac{6}{e^{3-x}}\right)^2 = \frac{36}{e^{6-2x}} = 36e^{2x-6}$

2. Integrate the simplified exponential expression

$\int 36e^{2x-6} \, dx = 36 \left( \frac{e^{2x-6}}{2} \right) + c$

3. Simplify the final answer

$= 18e^{2x-6} + c$

1. Find a linear factor of the cubic polynomial by trial and error.

Let $f(x) = 2x^3 + 3x^2 + 2x + 8$.
Try $x = -2$:
$f(-2) = 2(-2)^3 + 3(-2)^2 + 2(-2) + 8$
$f(-2) = 2(-8) + 3(4) - 4 + 8 = -16 + 12 - 4 + 8 = 0$
By the Factor Theorem, $(x + 2)$ is a factor.

2. Factorise the cubic polynomial into a linear and a quadratic factor.

$2x^3 + 3x^2 + 2x + 8 = (x + 2)(ax^2 + bx + c)$
By comparing coefficients:
$x^3$ term: $a = 2$
Constant term: $2c = 8 \implies c = 4$
$x^2$ term: $2a + b = 3 \implies 2(2) + b = 3 \implies b = -1$
So, $2x^3 + 3x^2 + 2x + 8 = (x + 2)(2x^2 - x + 4)$.

3. Evaluate the discriminant of the quadratic factor to show it has no real roots.

For the quadratic factor $2x^2 - x + 4 = 0$:
Discriminant $\Delta = b^2 - 4ac = (-1)^2 - 4(2)(4)$
$\Delta = 1 - 32 = -31$
Since $\Delta < 0$, the equation $2x^2 - x + 4 = 0$ has no real roots.

4. Conclude the explanation and state the single real root.

Since the quadratic factor yields no real roots, the only real root comes from the linear factor $x + 2 = 0$.
Therefore, the equation has only 1 real root.
This root is $x = -2$.