The equation of a polynomial is given by $\mathrm{p}(x) = 4x^3 + x - 5$.

1. Apply the Remainder Theorem by substituting $x = \frac{1}{2}$ into $\mathrm{p}(x)$

Remainder $= \mathrm{p}\left(\frac{1}{2}\right)$
$= 4\left(\frac{1}{2}\right)^3 + \left(\frac{1}{2}\right) - 5$

2. Evaluate the expression to find the remainder

$= 4\left(\frac{1}{8}\right) + \frac{1}{2} - 5$
$= \frac{1}{2} + \frac{1}{2} - 5$
$= 1 - 5$
$= -4$

1. Find a linear factor of $\mathrm{p}(x)$ using the Factor Theorem

Let $\mathrm{p}(x) = 4x^3 + x - 5$.
By trial and error,
$\mathrm{p}(1) = 4(1)^3 + 1 - 5 = 0$.
Therefore, $(x - 1)$ is a factor of $\mathrm{p}(x)$.

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

$4x^3 + x - 5 = (x - 1)(Ax^2 + Bx + C)$
By comparing coefficients:
$A = 4$
$C = 5$
Comparing coefficients of $x^2$:
$B - A = 0 \implies B = 4$
So, $\mathrm{p}(x) = (x - 1)(4x^2 + 4x + 5)$.

3. Evaluate the discriminant of the quadratic factor

For the quadratic equation $4x^2 + 4x + 5 = 0$,
Discriminant, $\Delta = b^2 - 4ac$
$\Delta = 4^2 - 4(4)(5)$
$\Delta = 16 - 80 = -64$

4. Conclude the number of real roots

Since $\Delta < 0$, the quadratic equation $4x^2 + 4x + 5 = 0$ has no real roots.
Thus, $x = 1$ is the only real root of $\mathrm{p}(x) = 0$.

1. Rewrite the exponential equation to reveal the polynomial structure

$2^{3y+2} + 2^y - 5 = 0$
$2^2(2^{3y}) + 2^y - 5 = 0$
$4(2^y)^3 + 2^y - 5 = 0$

2. Use the substitution $x = 2^y$ and the result from part (ii)

Let $x = 2^y$. The equation becomes:
$4x^3 + x - 5 = 0$
This is $\mathrm{p}(x) = 0$. From part (ii), the only real root is $x = 1$.

3. Solve for $y$

$2^y = 1$
$y = 0$