Given that $x^2 + 2x - 3$ is a factor of $\text{f}(x)$, where $\text{f}(x) = x^4 + 4x^3 + 2x^2 + ax - b$.

1. Factorise the given quadratic factor to find the linear factors.

Let $g(x) = x^2 + 2x - 3$.
$g(x) = (x - 1)(x + 3)$
Since $g(x)$ is a factor of $\text{f}(x)$, both $(x - 1)$ and $(x + 3)$ are factors of $\text{f}(x)$.

2. Apply the Factor Theorem for the first linear factor.

By the Factor Theorem, $\text{f}(1) = 0$.
$(1)^4 + 4(1)^3 + 2(1)^2 + a(1) - b = 0$
$1 + 4 + 2 + a - b = 0$
$a - b = -7 \quad \text{--- (1)}$

3. Apply the Factor Theorem for the second linear factor.

By the Factor Theorem, $\text{f}(-3) = 0$.
$(-3)^4 + 4(-3)^3 + 2(-3)^2 + a(-3) - b = 0$
$81 - 108 + 18 - 3a - b = 0$
$-9 - 3a - b = 0$
$3a + b = -9 \quad \text{--- (2)}$

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

Adding equation (1) and equation (2):
$(a - b) + (3a + b) = -7 + (-9)$
$4a = -16$
$a = -4$

Substitute $a = -4$ into equation (1):
$-4 - b = -7$
$b = 3$

1. Set up an identity to find the other quadratic factor using the values of $a$ and $b$ from part (a)

From part (a), $a = -4$ and $b = 3$, so $\text{f}(x) = x^4 + 4x^3 + 2x^2 - 4x - 3$.
Let the other quadratic factor be $x^2 + px + q$.
$x^4 + 4x^3 + 2x^2 - 4x - 3 = (x^2 + 2x - 3)(x^2 + px + q)$

2. Compare coefficients to find $p$ and $q$

Comparing the constant terms:
$-3q = -3 \implies q = 1$
Comparing the coefficients of $x^3$:
$p + 2 = 4 \implies p = 2$
Thus, the other quadratic factor is $x^2 + 2x + 1$.