Day 1 - Determine how many positive and negative solutions there are according to Descartes Theorem. Determine all the possible zeros according to Rational Roots Theorem. \( \begin{array}{ll}\text { 1. } x^{3}-3 x+28 & \text { 2. } x^{3}-4 x^{-2}+x-10 \\ \text { 3. } 2 x^{4}+6 x^{3}-7 x+9 & \text { 4. } 2 x^{3}+x^{2}-x-18 \\ \text { 5. } 4 x^{5}+3 x^{3}-2 x-14 & \text { 6. } 3 x^{4}+5 x^{3}-3 x+42 \\ \text { 7. } 8 x^{4}+4 x^{3}-10 x+15 & \text { 8. } 6 x^{3}-3 x^{2}+12 \\ \text { 9. } x^{3}-12 x^{2}+35 x-24 & \text { 10. } x^{3}-5 x^{2}-22 x+56\end{array} \)

1. For the polynomial \( x^{3}-3x+28 \): The signs of the coefficients are +, -, +. There is 1 positive solution. For negative solutions, substituting \( x \) with \( -x \) gives \( -x^{3}+3x+28 \), which has coefficients -,+,+, resulting in 2 sign changes, indicating 2 negative solutions. 2. The possible rational roots for \( x^{3}-4x^{-2}+x-10 \): The Rational Roots Theorem indicates testing factors of the constant term (-10) over the leading coefficient (1), thus giving potential rational roots of \( \pm 1, \pm 2, \pm 5, \pm 10 \). 3. For the polynomial \( 2x^{4}+6x^{3}-7x+9 \): The signs of the coefficients are +, +, -, +. There are 1 positive solution. For the negative solutions, substituting \( x \) with \( -x \) gives \( 2(-x)^{4}+6(-x)^{3}-7(-x)+9 \), resulting in +, -, +, +, giving 1 negative solution. 4. The possible rational roots for \( 2x^{4}+6x^{3}-7x+9 \): The Rational Roots Theorem presents possible rational roots as \( \pm 1, \pm 3, \pm 9, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{9}{2} \). 5. Analyzing \( 4x^{5}+3x^{3}-2x-14 \): The coefficient signs are +, +, -, -, which indicates 2 positive solutions. For negative solutions via \( -x \), it becomes -,+,+,- giving 1 negative solution. 6. Possible rational roots for \( 4x^{5}+3x^{3}-2x-14 \): Consider \( \pm 1, \pm 2, \pm 7, \pm 14, \pm \frac{1}{4}, \pm \frac{7}{4} \). 7. In polynomial \( 8x^{4}+4x^{3}-10x+15 \): The signs are +, +, -, +, resulting in 2 positive solutions. With \( -x \), it becomes -,+,+,- which has 1 negative solution. 8. Possible rational roots for \( 8x^{4}+4x^{3}-10x+15 \): Potential roots can be \( \pm 1, \pm 3, \pm 5, \pm 15, \pm \frac{1}{8}, \pm \frac{3}{8}, \pm \frac{5}{8}, \pm \frac{15}{8} \). 9. Evaluating \( x^{3}-12x^{2}+35x-24 \): The signs are +, -, +, -. Thus, it has 2 positive solutions. For negatives, transforming via \( -x \) gives +, -,-,+, leading to 1 negative solution. 10. Possible rational roots for \( x^{3}-12x^{2}+35x-24 \): Possible roots are factors of -24 over 1, hence \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24 \). 11. Considering \( x^{3}-5x^{2}-22x+56 \): Signs are +, -, -, + indicating 1 positive and checking \( -x \) results in positive, giving 1 negative solution. 12. For rational roots from the last equation, possibilities are \( \pm 1, \pm 2, \pm 4, \pm 7, \pm 8, \pm 14, \pm 28, \pm 56 \). Now you have a well-rounded view on positive and negative solutions according to Descartes Theorem and potential zeros suggested by the Rational Roots Theorem! Happy solving!