1. **Problem 1:** Find the remainder when $P(x) = 4x^3 - 4x + 1$ is divided by $2x - 3$. 2. Use the Remainder Theorem: The remainder when a polynomial $P(x)$ is divided by a linear divisor $ax - b$ is $P\left(\frac{b}{a}\right)$. 3. Here, divisor is $2x - 3$, so $a=2$, $b=3$. Evaluate $P\left(\frac{3}{2}\right)$: $$P\left(\frac{3}{2}\right) = 4\left(\frac{3}{2}\right)^3 - 4\left(\frac{3}{2}\right) + 1$$ 4. Calculate powers and multiply: $$= 4 \times \frac{27}{8} - 4 \times \frac{3}{2} + 1 = \frac{108}{8} - 6 + 1 = \frac{27}{2} - 5$$ 5. Convert $5$ to halves: $5 = \frac{10}{2}$, so: $$\frac{27}{2} - \frac{10}{2} = \frac{17}{2}$$ 6. So, the remainder is $\frac{17}{2}$. 7. **Problem 2:** Which binomial is a factor of $x^3 - 6x^2 + 11x - 6$? 8. Use the Factor Theorem: If $(x - c)$ is a factor, then $P(c) = 0$. 9. Test each candidate: - For $x + 1$, test $x = -1$: $$(-1)^3 - 6(-1)^2 + 11(-1) - 6 = -1 - 6 - 11 - 6 = -24 \neq 0$$ - For $x - 1$, test $x = 1$: $$1 - 6 + 11 - 6 = 0$$ 10. Since $P(1) = 0$, $x - 1$ is a factor. 11. **Problem 3:** Which set of values for $x$ should be tested to find possible zeros of $2x^3 - 3x^2 + 3x - 10$? 12. Use Rational Root Theorem: Possible zeros are $\pm$ factors of constant term over factors of leading coefficient. 13. Constant term is $-10$, factors: $\pm1, \pm2, \pm5, \pm10$. 14. Leading coefficient is $2$, factors: $\pm1, \pm2$. 15. Possible zeros: $$\pm1, \pm\frac{1}{2}, \pm2, \pm5, \pm\frac{5}{2}, \pm10$$ **Final answers:** - Remainder when dividing $P(x)$ by $2x - 3$ is $\frac{17}{2}$. - The binomial factor of $x^3 - 6x^2 + 11x - 6$ is $x - 1$. - Possible zeros to test for $2x^3 - 3x^2 + 3x - 10$ are $\pm1, \pm\frac{1}{2}, \pm2, \pm5, \pm\frac{5}{2}, \pm10$.