1. **State the problem:** We need to find the expression for $h(x+1)$ given the function $h(x) = 3x^4 + x^3 - 4x^2 + x - 2$. 2. **Formula and approach:** To find $h(x+1)$, substitute $x+1$ into every $x$ in the function: $$h(x+1) = 3(x+1)^4 + (x+1)^3 - 4(x+1)^2 + (x+1) - 2$$ 3. **Expand each term:** - Expand $(x+1)^4$ using the binomial theorem: $$ (x+1)^4 = x^4 + 4x^3 + 6x^2 + 4x + 1 $$ - Expand $(x+1)^3$: $$ (x+1)^3 = x^3 + 3x^2 + 3x + 1 $$ - Expand $(x+1)^2$: $$ (x+1)^2 = x^2 + 2x + 1 $$ 4. **Substitute expansions back:** $$h(x+1) = 3(x^4 + 4x^3 + 6x^2 + 4x + 1) + (x^3 + 3x^2 + 3x + 1) - 4(x^2 + 2x + 1) + (x + 1) - 2$$ 5. **Distribute coefficients:** $$= 3x^4 + 12x^3 + 18x^2 + 12x + 3 + x^3 + 3x^2 + 3x + 1 - 4x^2 - 8x - 4 + x + 1 - 2$$ 6. **Combine like terms:** - $x^4$ terms: $3x^4$ - $x^3$ terms: $12x^3 + x^3 = 13x^3$ - $x^2$ terms: $18x^2 + 3x^2 - 4x^2 = 17x^2$ - $x$ terms: $12x + 3x - 8x + x = 8x$ - Constants: $3 + 1 - 4 + 1 - 2 = -1$ 7. **Final expression:** $$h(x+1) = 3x^4 + 13x^3 + 17x^2 + 8x - 1$$ This is the expanded form of $h(x+1)$ after substitution and simplification.