1. **State the problem:** We are given the equation $$6x^3 + 5x^2 - 4x + 22 = (A + B + C)x^3 + A(x - 1)^2 + B(x + 2) - C$$ and need to find the values of constants $A$, $B$, and $C$ such that the equality holds for all values of $x$. 2. **Expand and simplify the right-hand side:** First, expand each term: - Expand $A(x - 1)^2 = A(x^2 - 2x + 1) = A x^2 - 2A x + A$ - Expand $B(x + 2) = B x + 2B$ So the right side becomes: $$(A + B + C) x^3 + A x^2 - 2 A x + A + B x + 2 B - C$$ 3. **Group like terms on the right side:** $$x^3: (A + B + C)$$ $$x^2: A$$ $$x: (-2 A + B)$$ $$ ext{constant}: (A + 2 B - C)$$ 4. **Set coefficients equal to those on the left side:** From the original polynomial $6x^3 + 5x^2 - 4x + 22$, equate coefficients: $$x^3: 6 = A + B + C$$ $$x^2: 5 = A$$ $$x: -4 = -2 A + B$$ $$ ext{constant}: 22 = A + 2 B - C$$ 5. **Substitute $A = 5$ into the other equations:** $$6 = 5 + B + C \implies B + C = 1$$ $$-4 = -2(5) + B \implies -4 = -10 + B \implies B = 6$$ $$22 = 5 + 2 B - C \implies 22 = 5 + 2(6) - C \implies 22 = 5 + 12 - C \implies 22 = 17 - C \implies C = -5$$ 6. **Check the first equation with found values:** $$B + C = 6 + (-5) = 1$$ which matches the equation. 7. **Final answer:** $$A = 5, \quad B = 6, \quad C = -5$$