Answer
$P(x)=\frac{-2h}{l^3}x^3+\frac{3h}{l^2}x^2$
Work Step by Step
$P(0)=0; \hspace{3mm}P(l)=h;\hspace{3mm}P'(0)=0;\hspace{3mm}P'(l)=0$
$P(x)=ax^3+bx^2+cx+d$
$P(0)=d=0$
$P(l)=al^3+bl^2+cl=h$
$P'(x)=3ax^2+2bx+c$
$P'(0)=c=0$
$P'(l)=3al^2+2bl+c=0$
$al^3+bl^2=h$
$a=\frac{h-bl^2}{l^3}$
$3al^2+2bl=0$
$3(\frac{h-bl^2}{l^3})l^2+2bl=0$
$lb=\frac{3h}{l}$
$b=\frac{3h}{l^2}$
$a=\frac{h-3h}{l^3}=\frac{-2h}{l^3}$
$P(x)=\frac{-2h}{l^3}x^3+\frac{3h}{l^2}x^2$
