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