1. **State the problem:** Expand the cubic polynomial $$f(x) = 2x^3 + 7x^2 + x - 6$$ in powers of $$(x-2)$$. 2. **Recall the binomial expansion formula:** To expand a function around $x=a$, we write it as a polynomial in $(x-a)$: $$f(x) = c_0 + c_1(x-a) + c_2(x-a)^2 + c_3(x-a)^3 + \cdots$$ where the coefficients $c_n$ can be found by substituting $x = (x-2) + 2$ and expanding. 3. **Rewrite $f(x)$ substituting $x = (x-2) + 2$:** $$f(x) = 2((x-2)+2)^3 + 7((x-2)+2)^2 + ((x-2)+2) - 6$$ 4. **Expand each term:** - Expand $((x-2)+2)^3$ using binomial expansion: $$((x-2)+2)^3 = (x-2)^3 + 3 \cdot 2 (x-2)^2 + 3 \cdot 2^2 (x-2) + 2^3 = (x-2)^3 + 6(x-2)^2 + 12(x-2) + 8$$ - Expand $((x-2)+2)^2$: $$((x-2)+2)^2 = (x-2)^2 + 2 \cdot 2 (x-2) + 2^2 = (x-2)^2 + 4(x-2) + 4$$ 5. **Substitute expansions back into $f(x)$:** $$f(x) = 2[(x-2)^3 + 6(x-2)^2 + 12(x-2) + 8] + 7[(x-2)^2 + 4(x-2) + 4] + (x-2) + 2 - 6$$ 6. **Distribute coefficients:** $$= 2(x-2)^3 + 12(x-2)^2 + 24(x-2) + 16 + 7(x-2)^2 + 28(x-2) + 28 + (x-2) + 2 - 6$$ 7. **Group like terms:** - Cubic term: $$2(x-2)^3$$ - Quadratic terms: $$12(x-2)^2 + 7(x-2)^2 = 19(x-2)^2$$ - Linear terms: $$24(x-2) + 28(x-2) + (x-2) = 53(x-2)$$ - Constants: $$16 + 28 + 2 - 6 = 40$$ 8. **Final expanded form:** $$\boxed{f(x) = 2(x-2)^3 + 19(x-2)^2 + 53(x-2) + 40}$$ This is the polynomial $f(x)$ expanded in powers of $(x-2)$.