Constants A, $B$ and $C$ satisfy $x^2+3x+5=A(x-1{)}^2+B(x-1)+C$ for every real $x$. Then $A+B+C$ equals:

Since the identity holds for every $x$, put $x=2$: the right side becomes $A+B+C$ and the left side becomes 4 + 6 + 5 = 15. (Putting $x=1$ isolates $C=9$, and $x=0$ gives the constant 5 — each answers a different question; 11 comes from taking $C=5$.) Comparing coefficients confirms it: $A=1,B-2A=3$ so $B=5$, and $A-B+C=5$ so $C=9$, and 1 + 5 + 9 = 15.