Polynomial Division Cef01C

1. **Problem 65:** Divide $c^3 - 3c^2 + 18c - 16$ by $c^2 + 3c - 2$ using long division. 2. **Step 1:** Write the division setup: $$\frac{c^3 - 3c^2 + 18c - 16}{c^2 + 3c - 2}$$ 3. **Step 2:** Divide the leading term $c^3$ by $c^2$ to get $c$. 4. **Step 3:** Multiply $c$ by the divisor: $$c(c^2 + 3c - 2) = c^3 + 3c^2 - 2c$$ 5. **Step 4:** Subtract this from the dividend: $$\begin{aligned} &(c^3 - 3c^2 + 18c - 16) - (c^3 + 3c^2 - 2c) \\ &= (c^3 - c^3) + (-3c^2 - 3c^2) + (18c + 2c) - 16 \\ &= 0 - 6c^2 + 20c - 16 \end{aligned}$$ 6. **Step 5:** Divide the new leading term $-6c^2$ by $c^2$ to get $-6$. 7. **Step 6:** Multiply $-6$ by the divisor: $$-6(c^2 + 3c - 2) = -6c^2 - 18c + 12$$ 8. **Step 7:** Subtract this from the previous remainder: $$\begin{aligned} &(-6c^2 + 20c - 16) - (-6c^2 - 18c + 12) \\ &= (-6c^2 + 6c^2) + (20c + 18c) + (-16 - 12) \\ &= 0 + 38c - 28 \end{aligned}$$ 9. **Step 8:** Since the degree of the remainder $38c - 28$ is less than the divisor degree, stop. 10. **Step 9:** Write the result: $$c - 6 + \frac{38c - 28}{c^2 + 3c - 2}$$ --- 11. **Problem 66:** Divide $x^4 - 2x^2 - x + 2$ by $x + 2$ using synthetic division. 12. **Step 1:** Set up synthetic division with root $-2$ (since divisor is $x + 2$): Coefficients: $1$ (for $x^4$), $0$ (for $x^3$), $-2$ (for $x^2$), $-1$ (for $x$), $2$ (constant). 13. **Step 2:** Perform synthetic division: $$\begin{array}{r|rrrrr} -2 & 1 & 0 & -2 & -1 & 2 \\ & & -2 & 4 & -4 & 10 \\ \hline & 1 & -2 & 2 & -5 & 12 \\ \end{array}$$ 14. **Step 3:** The quotient is: $$x^3 - 2x^2 + 2x - 5$$ 15. **Step 4:** The remainder is $12$, so the full answer is: $$x^3 - 2x^2 + 2x - 5 + \frac{12}{x + 2}$$