Given that $(x+2)$ and $(2x+3)$ are factors of $f(x) = 6x^3 + px^2 + 4x - q$, we need to find the values of $p$ and $q$, and the roots of $f(x)$.

1. Problem Description

Given that $(x+2)$ and $(2x+3)$ are factors of $f(x) = 6x^3 + px^2 + 4x - q$, we need to find the values of $p$ and $q$, and the roots of $f(x)$.

2. Solution Steps

(a) Since $(x+2)$ and $(2x+3)$ are factors of $f(x)$, we can use the factor theorem. This means that $f(-2) = 0$ and $f(-\frac{3}{2}) = 0$.

$f(-2) = 6(-2)^3 + p(-2)^2 + 4(-2) - q = 0$

$-48 + 4p - 8 - q = 0$

$4p - q = 56$ (Equation 1)

$f(-\frac{3}{2}) = 6(-\frac{3}{2})^3 + p(-\frac{3}{2})^2 + 4(-\frac{3}{2}) - q = 0$

$6(-\frac{27}{8}) + p(\frac{9}{4}) - 6 - q = 0$

$-\frac{81}{4} + \frac{9}{4}p - 6 - q = 0$

Multiply by 4 to get rid of fractions:

$-81 + 9p - 24 - 4q = 0$

$9p - 4q = 105$ (Equation 2)

Now we solve the system of equations:

$4p - q = 56$

$9p - 4q = 105$

From Equation 1, $q = 4p - 56$. Substitute this into Equation 2:

$9p - 4(4p - 56) = 105$

$9p - 16p + 224 = 105$

$-7p = -119$

$p = 17$

Now substitute $p=17$ into $q = 4p - 56$:

$q = 4(17) - 56 = 68 - 56 = 12$

So, $p = 17$ and $q = 12$.

(b) Now we know $f(x) = 6x^3 + 17x^2 + 4x - 12$. We know that $(x+2)$ and $(2x+3)$ are factors. Since $f(x)$ is a cubic polynomial, it has three roots. We already know two roots are $x = -2$ and $x = -\frac{3}{2}$. We want to find the third root.

Since $(x+2)$ and $(2x+3)$ are factors, $(x+2)(2x+3) = 2x^2 + 7x + 6$ is a factor of $f(x)$.

We can perform polynomial division to find the other factor.

$(6x^3 + 17x^2 + 4x - 12) / (2x^2 + 7x + 6) = 3x - 2$

So $f(x) = (x+2)(2x+3)(3x-2)$.

Thus, the roots are $x = -2$, $x = -\frac{3}{2}$, and $x = \frac{2}{3}$.

3. Final Answer

(a) $p = 17$, $q = 12$

(b) The roots of $f(x)$ are $x = -2$, $x = -\frac{3}{2}$, and $x = \frac{2}{3}$.