多项式 $\mathrm{f}(x)$ 满足 $x^4$ 的系数为 $2$。方程 $\mathrm{f}(x) = 0$ 的根为 $\frac{3}{2}$ 和 $-1$。$\mathrm{f}(x)$ 除以 $x-1$ 的余数为 $-8$，除以 $x+2$ 的余数为 $28$。

1. Set up the general form of the polynomial using the given roots and leading coefficient.

Since the roots are $\frac{3}{2}$ and $-1$, $(2x-3)$ and $(x+1)$ are factors of $f(x)$.
Given that the coefficient of $x^4$ is $2$, we can assume $f(x)$ is a quartic polynomial of the form:
$f(x) = (2x-3)(x+1)(x^2 + px + q)$

2. Use the Remainder Theorem to form equations for $p$ and $q$.

By the Remainder Theorem, $f(1) = -8$ and $f(-2) = 28$.
For $f(1) = -8$:
$(2(1)-3)(1+1)(1^2 + p(1) + q) = -8$
$(-1)(2)(1 + p + q) = -8$
$-2(1 + p + q) = -8$
$1 + p + q = 4$
$p + q = 3 \quad \text{--- (1)}$

For $f(-2) = 28$:
$(2(-2)-3)(-2+1)((-2)^2 + p(-2) + q) = 28$
$(-7)(-1)(4 - 2p + q) = 28$
$7(4 - 2p + q) = 28$
$4 - 2p + q = 4$
$-2p + q = 0 \implies q = 2p \quad \text{--- (2)}$

3. Solve the simultaneous equations to find $p$ and $q$.

Substitute (2) into (1):
$p + 2p = 3$
$3p = 3 \implies p = 1$

Substitute $p = 1$ into (2):
$q = 2(1) = 2$

4. Write the final expression for $f(x)$.

Substitute $p$ and $q$ back into the expression for $f(x)$:
$f(x) = (2x-3)(x+1)(x^2 + x + 2)$
$f(x) = (2x^2 - x - 3)(x^2 + x + 2)$
$f(x) = 2x^4 + 2x^3 + 4x^2 - x^3 - x^2 - 2x - 3x^2 - 3x - 6$
$f(x) = 2x^4 + x^3 - 5x - 6$

1. Use the expression for $f(x)$ from part (a) to identify the remaining quadratic factor.

From part (a), we found that $\mathrm{f}(x) = (2x-3)(x+1)(x^2+x+2)$.
For $\mathrm{f}(x) = 0$, the roots are $x = \frac{3}{2}$, $x = -1$, or the roots of $x^2+x+2 = 0$.

2. Evaluate the discriminant of the quadratic factor to check for real roots.

For $x^2+x+2 = 0$, the discriminant is:
$b^2 - 4ac = (1)^2 - 4(1)(2) = 1 - 8 = -7$
Since the discriminant is less than $0$, $x^2+x+2 = 0$ has no real roots.

3. State the total number of real roots.

Therefore, the only real roots are $\frac{3}{2}$ and $-1$.
There are exactly 2 real roots.