Question: Divide using long division. State the quotient, $q(x)$, and the remainder, $r(x)$. $$(x^3 + 2x^2 - 8x + 5) \div (x - 1)$$ $$(x^3 + 2x^2 - 8x + 5) \div (x - 1) = \square + \square/(x - 1)$$ (Simplify your answers. Do not factor.)
1. **State the problem:** Divide the polynomial $$x^3 + 2x^2 - 8x + 5$$ by the binomial $$x - 1$$ using long division.
2. **Recall the formula:** When dividing polynomials, the division algorithm states:
$$\text{Dividend} = (\text{Divisor}) \times (\text{Quotient}) + \text{Remainder}$$
3. **Set up the long division:** Divide $$x^3$$ by $$x$$ to get the first term of the quotient: $$x^2$$.
4. **Multiply:** Multiply $$x^2$$ by the divisor $$x - 1$$:
$$x^2 \times (x - 1) = x^3 - x^2$$.
5. **Subtract:**
$$\begin{aligned}
(x^3 + 2x^2 - 8x + 5) - (x^3 - x^2) &= (x^3 - x^3) + (2x^2 - (-x^2)) - 8x + 5 \\
&= 0 + 3x^2 - 8x + 5
\end{aligned}$$
6. **Bring down the next term:** The new dividend is $$3x^2 - 8x + 5$$.
7. **Divide:** Divide $$3x^2$$ by $$x$$ to get $$3x$$.
8. **Multiply:** Multiply $$3x$$ by $$x - 1$$:
$$3x \times (x - 1) = 3x^2 - 3x$$.
9. **Subtract:**
$$\begin{aligned}
(3x^2 - 8x + 5) - (3x^2 - 3x) &= (3x^2 - 3x^2) + (-8x - (-3x)) + 5 \\
&= 0 - 5x + 5
\end{aligned}$$
10. **Bring down the next term:** The new dividend is $$-5x + 5$$.
11. **Divide:** Divide $$-5x$$ by $$x$$ to get $$-5$$.
12. **Multiply:** Multiply $$-5$$ by $$x - 1$$:
$$-5 \times (x - 1) = -5x + 5$$.
13. **Subtract:**
$$(-5x + 5) - (-5x + 5) = 0$$.
14. **Conclusion:** The quotient is $$q(x) = x^2 + 3x - 5$$ and the remainder is $$r(x) = 0$$.
15. **Final answer:**
$$\frac{x^3 + 2x^2 - 8x + 5}{x - 1} = x^2 + 3x - 5 + \frac{0}{x - 1} = x^2 + 3x - 5$$