Since (x - 1), is a factor, when the polynomial is divided by (x - 1), the remainder = zero

\(\therefore (x - 1) = 0\)

x = 1

Substitute in the polynomial the value x = 1

= \(p(1)^3 + q(1)^2 + 11(1) - 6 = 0\)

p + q + 5 = 0 .....(i)

Also since x - 3 is a factor, \(\therefore\) x - 3 = 0

x = 3

Substitute \(p(3)^3 + q(3)^2 + 11(3) - 6 = 0\)

27p + 9q = -27 ......(2)

Combine eqns. (i) and (ii)

Multiply equation (i) by 9 to eliminate q

9p + 9q = -45

Subtract (ii) from (i), \(18p = 18\)

\(\therefore\) p = 1

Put p = 1 in (i), 

\(1 + q = -5 \implies q = -6\)

\((p, q) = (1, -6)\)