Use a remainder and a stationary point as constraints

Let $p(x)=ax^3+bx^2+cx+d$. The remainder theorem gives $p(2)=52$, so $8a+4b+2c+d=52$. The given point gives $3.375a+2.25b+1.5c+d=32.375$. The stationary point supplies both $-a+b-c+d=-2$ and $p^\prime(-1)=3a-2b+c=0$.

The source GC solution is $a=1,b=6,c=9,d=2$, hence $p(x)=x^3+6x^2+9x+2$. It gives $p(2)=52$, $p(1.5)=32.375$, $p(-1)=-2$, $p^\prime(-1)=0$. The source working’s label “$p(2)=54$” is a typo: its actual coefficient equation, prompt and final polynomial all use 52. Original 54 retained in the native source.