Problem Analysis: The question asks for the relationship between $p$ and $q$ given that $(4y - 1)$ and $(y + 4)$ are factors of the quadratic polynomial $py^2 + 15y - q$. We can use the Factor Theorem, which states that if $(x - a)$ is a factor of a polynomial $P(x)$, then $P(a) = 0$.

Applying the Factor Theorem

Since $(4y - 1)$ is a factor, $y = \frac{1}{4}$ must be a root of the polynomial.

• Substitute $y = \frac{1}{4}$ into the polynomial: $p\left(\frac{1}{4}\right)^2 + 15\left(\frac{1}{4}\right) - q = 0$ $p\left(\frac{1}{16}\right) + \frac{15}{4} - q = 0$ Multiply by 16 to clear fractions: $p + 4 \times 15 - 16q = 0$ $p + 60 - 16q = 0$ $p - 16q = -60$ (Equation 1)

Since $(y + 4)$ is a factor, $y = -4$ must be a root of the polynomial.

• Substitute $y = -4$ into the polynomial: $p(-4)^2 + 15(-4) - q = 0$ $p(16) - 60 - q = 0$ $16p - q = 60$ (Equation 2)

Solving for p and q

We now have a system of two linear equations with two variables ($p$ and $q$): From Equation 2, we can express $q$ in terms of $p$: $q = 16p - 60$ Substitute this expression for $q$ into Equation 1: $p - 16(16p - 60) = -60$ $p - 256p + 960 = -60$ $-255p = -60 - 960$ $-255p = -1020$ $p = \frac{-1020}{-255}$ $p = 4$ Now substitute the value of $p$ back into the expression for $q$: $q = 16p - 60$ $q = 16(4) - 60$ $q = 64 - 60$ $q = 4$

1. $p - 16q = -60$
2. $16p - q = 60$

Determining the Relationship

We found $p = 4$ and $q = 4$. Therefore, the relationship between $p$ and $q$ is:

$p = q$