Polynomial Expansion 17Ef12

1. **State the problem:** We want to express the polynomial $X^2+2x-1$ in terms of the given expression involving $A$, $B$, and $C$: $$X^2+2x-1=A(x-1)^2(x+2)+B(x-1)(x+2)+C(x-1)^2(x-1)$$ 2. **Expand each term:** - Expand $(x-1)^2 = (x-1)(x-1) = x^2 - 2x + 1$ - Then: - $A(x-1)^2(x+2) = A(x^2 - 2x + 1)(x+2)$ - $B(x-1)(x+2)$ is already factored - $C(x-1)^2(x-1) = C(x^2 - 2x + 1)(x-1)$ 3. **Expand $A$ term:** $$A(x^2 - 2x + 1)(x+2) = A[(x^2)(x+2) - 2x(x+2) + 1(x+2)]$$ $$= A[x^3 + 2x^2 - 2x^2 - 4x + x + 2] = A[x^3 + (2x^2 - 2x^2) + (-4x + x) + 2] = A[x^3 - 3x + 2]$$ 4. **Expand $B$ term:** $$B(x-1)(x+2) = B(x^2 + 2x - x - 2) = B(x^2 + x - 2)$$ 5. **Expand $C$ term:** $$C(x^2 - 2x + 1)(x-1) = C[(x^2)(x-1) - 2x(x-1) + 1(x-1)]$$ $$= C[x^3 - x^2 - 2x^2 + 2x + x - 1] = C[x^3 - 3x^2 + 3x - 1]$$ 6. **Sum all terms:** $$A[x^3 - 3x + 2] + B[x^2 + x - 2] + C[x^3 - 3x^2 + 3x - 1]$$ $$= (A + C)x^3 + (B - 3C)x^2 + (-3A + B + 3C)x + (2A - 2B - C)$$ 7. **Equate coefficients with $X^2 + 2x - 1$:** The polynomial on the left is $0x^3 + 1x^2 + 2x - 1$. So, - Coefficient of $x^3$: $A + C = 0$ - Coefficient of $x^2$: $B - 3C = 1$ - Coefficient of $x$: $-3A + B + 3C = 2$ - Constant term: $2A - 2B - C = -1$ 8. **Solve the system:** From $A + C = 0$, we get $C = -A$. Substitute $C = -A$ into other equations: - $B - 3(-A) = 1 ightarrow B + 3A = 1$ - $-3A + B + 3(-A) = 2 ightarrow -3A + B - 3A = 2 ightarrow B - 6A = 2$ - $2A - 2B - (-A) = -1 ightarrow 2A - 2B + A = -1 ightarrow 3A - 2B = -1$ From $B + 3A = 1$ and $B - 6A = 2$: Subtract second from first: $B + 3A - (B - 6A) = 1 - 2 ightarrow 3A + 6A = -1 ightarrow 9A = -1 ightarrow A = -\frac{1}{9}$ Then $B = 1 - 3A = 1 - 3(-\frac{1}{9}) = 1 + \frac{1}{3} = \frac{4}{3}$ From $3A - 2B = -1$: $3(-\frac{1}{9}) - 2(\frac{4}{3}) = -\frac{1}{3} - \frac{8}{3} = -3$ which does not equal $-1$. This indicates a contradiction, so re-check calculations. 9. **Re-check constant term equation:** $2A - 2B - C = -1$ Substitute $C = -A$: $2A - 2B - (-A) = 2A - 2B + A = 3A - 2B = -1$ Using $B = 1 - 3A$: $3A - 2(1 - 3A) = -1$ $3A - 2 + 6A = -1$ $9A - 2 = -1$ $9A = 1$ $A = \frac{1}{9}$ Now recalculate $B$: $B = 1 - 3A = 1 - 3(\frac{1}{9}) = 1 - \frac{1}{3} = \frac{2}{3}$ Check $B - 6A = 2$: $\frac{2}{3} - 6(\frac{1}{9}) = \frac{2}{3} - \frac{2}{3} = 0 \neq 2$ There is a mismatch, so re-examine the $x$ coefficient equation: $-3A + B + 3C = 2$ Substitute $C = -A$: $-3A + B - 3A = 2$ $B - 6A = 2$ Using $B = 1 - 3A$: $1 - 3A - 6A = 2$ $1 - 9A = 2$ $-9A = 1$ $A = -\frac{1}{9}$ Then $B = 1 - 3(-\frac{1}{9}) = 1 + \frac{1}{3} = \frac{4}{3}$ Check constant term: $3A - 2B = -1$ $3(-\frac{1}{9}) - 2(\frac{4}{3}) = -\frac{1}{3} - \frac{8}{3} = -3 \neq -1$ This inconsistency suggests the original problem might have a typo or the last term $C(x-1)^2(x-1)$ should be $C(x-1)^2(x+1)$ or similar. 10. **Conclusion:** The system is inconsistent with the given expression. Please verify the problem statement for correctness.