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, \(\therefore\)Â 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), \(18p = 18\)
\(\therefore\)Â p = 1
Put p = 1 in (i),Â
\(1 + q = -5 \implies q = -6\)
\((p, q) = (1, -6)\)