1. **State the problem:** We want to express the polynomial $$2x^3 - 8x^2 + 3x - 4$$ in the form $$a(x - 1)^3 + b(x - 1)^2 + c(x - 1) + d$$ for all values of $$x$$ and find the constants $$a, b, c, d$$. 2. **Recall the binomial expansion:** $$(x - 1)^3 = x^3 - 3x^2 + 3x - 1$$ $$(x - 1)^2 = x^2 - 2x + 1$$ 3. **Expand the right-hand side:** $$a(x - 1)^3 + b(x - 1)^2 + c(x - 1) + d = a(x^3 - 3x^2 + 3x - 1) + b(x^2 - 2x + 1) + c(x - 1) + d$$ 4. **Distribute coefficients:** $$= a x^3 - 3a x^2 + 3a x - a + b x^2 - 2b x + b + c x - c + d$$ 5. **Group like terms:** $$= a x^3 + (-3a + b) x^2 + (3a - 2b + c) x + (-a + b - c + d)$$ 6. **Match coefficients with the original polynomial:** $$2x^3 - 8x^2 + 3x - 4 = a x^3 + (-3a + b) x^2 + (3a - 2b + c) x + (-a + b - c + d)$$ Equate coefficients: - For $$x^3$$: $$2 = a$$ - For $$x^2$$: $$-8 = -3a + b$$ - For $$x$$: $$3 = 3a - 2b + c$$ - For constant term: $$-4 = -a + b - c + d$$ 7. **Substitute $$a = 2$$ into the other equations:** - $$-8 = -3(2) + b \Rightarrow -8 = -6 + b \Rightarrow b = -2$$ - $$3 = 3(2) - 2(-2) + c \Rightarrow 3 = 6 + 4 + c \Rightarrow c = 3 - 10 = -7$$ - $$-4 = -2 + (-2) - (-7) + d \Rightarrow -4 = -2 - 2 + 7 + d \Rightarrow -4 = 3 + d \Rightarrow d = -7$$ 8. **Final answer:** $$a = 2, b = -2, c = -7, d = -7$$