56. What are the solutions to the polynomial: \( p(x)=(x-5)(3 x+5)\left(x^{2}-7 x+15\right) \)

To find the solutions to the polynomial \( p(x)=(x-5)(3 x+5)\left(x^{2}-7 x+15\right) \), we can break it down into its factors: 1. The linear factor \( x - 5 = 0 \) gives us the solution \( x = 5 \). 2. The linear factor \( 3x + 5 = 0 \) leads to \( 3x = -5 \) or \( x = -\frac{5}{3} \). 3. The quadratic factor \( x^{2}-7x+15 = 0 \) can be solved using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \). For this polynomial, \( a = 1, b = -7, c = 15 \), which gives us \( x = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot 15}}{2 \cdot 1} = \frac{7 \pm \sqrt{49 - 60}}{2} = \frac{7 \pm \sqrt{-11}}{2} \). Thus, this gives two complex solutions: \( x = \frac{7}{2} \pm \frac{\sqrt{11}}{2} i \). In summary, the solutions to \( p(x) \) are: 1. \( x = 5 \) 2. \( x = -\frac{5}{3} \) 3. \( x = \frac{7}{2} + \frac{\sqrt{11}}{2} i \) 4. \( x = \frac{7}{2} - \frac{\sqrt{11}}{2} i \)