p ' (x) = k(x – 1)(x – 3)  =  k(x² – 4x + 3)

p'(x) = $\mathrm{k}(\frac{{\mathrm{x}}^3}{3}-\frac{4{\mathrm{x}}^2}{2}+3\mathrm{x})+\mathrm{\lambda }$

p(1) = 6 

 $\therefore \mathrm{k}(\frac{1}{3}-2+3)+\mathrm{\lambda }=6\Rightarrow \frac{4\mathrm{k}}{3}+\mathrm{\lambda }=6$

p(3) = 2

$\therefore$  k(9 – 18 + 9) + $\mathrm{\lambda }$ = 2 $\Rightarrow \mathrm{\lambda }$ = 2

 $\therefore \frac{4\mathrm{k}}{3}=4\Rightarrow \mathrm{k}=3$

 $\therefore$        p'(x) = 3(x² – 4x + 3) $\Rightarrow$ p'(0) = 9