Given that $\text{p}(x) = (x+1)^3 + 5(x-1)^2 + ax + b$ has a factor of $x-1$ and leaves a remainder of 21 when divided by $x-2$, find the value of $a$ and of $b$.

1. Apply the Factor Theorem to form the first equation

Since $x-1$ is a factor, $\text{p}(1) = 0$.
$(1+1)^3 + 5(1-1)^2 + a(1) + b = 0$
$2^3 + 0 + a + b = 0$
$8 + a + b = 0$
$a + b = -8 \quad \text{--- (1)}$

2. Apply the Remainder Theorem to form the second equation

Since the remainder is 21 when divided by $x-2$, $\text{p}(2) = 21$.
$(2+1)^3 + 5(2-1)^2 + a(2) + b = 21$
$3^3 + 5(1)^2 + 2a + b = 21$
$27 + 5 + 2a + b = 21$
$32 + 2a + b = 21$
$2a + b = -11 \quad \text{--- (2)}$

3. Solve the simultaneous equations to find $a$ and $b$

Subtracting equation (1) from equation (2):
$(2a + b) - (a + b) = -11 - (-8)$
$a = -3$

Substitute $a = -3$ into equation (1):
$-3 + b = -8$
$b = -5$