(a) Find the rational zeros and then the other zeros of the polynomial function \( f(x)=3 x^{3}+5 x^{2}+75 x+125 \), that is, solve \( f(x)=0 \). (b) Factor \( f(x) \) into linear factors. (a) Select the correct choice below and fill in any answer box(es) within your choice. (Type an exact answer, using radicals and \( i \) as needed. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.) B. There is only one rational zero, \( \square \), and the other zeros are \( \square \). C. There are no rational zeros. The other zeros are Beros and they are
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To tackle the polynomial \( f(x) = 3x^3 + 5x^2 + 75x + 125 \), we can start by using the Rational Root Theorem to identify possible rational zeros. The factors of the constant term (125) and the leading coefficient (3) suggest checking values like \( \pm 1, \pm 5, \pm 25, \pm 125, \pm \frac{1}{3}, \pm \frac{5}{3}, \pm \frac{25}{3}, \pm \frac{125}{3} \). Testing \( x = -5 \): \[ f(-5) = 3(-5)^3 + 5(-5)^2 + 75(-5) + 125 = -375 + 125 - 375 + 125 = -500 \neq 0. \] After testing a few, we find \( x = -5 \) is indeed a zero. We can then divide \( f(x) \) by \( x + 5 \) using synthetic division, leading us to: \[ f(x) = (x + 5)(3x^2 - 10x + 25). \] Next, we solve the quadratic \( 3x^2 - 10x + 25 = 0 \) using the quadratic formula. The discriminant \( (-10)^2 - 4(3)(25) = 100 - 300 = -200 \) is negative, indicating two complex zeros: \[ x = \frac{10 \pm i\sqrt{200}}{6} = \frac{10 \pm 10i\sqrt{2}}{6} = \frac{5 \pm 5i\sqrt{2}}{3}. \] So we have found the rational zero \( -5 \) and the complex zeros \( \frac{5 + 5i\sqrt{2}}{3}, \frac{5 - 5i\sqrt{2}}{3} \). Now, factor \( f(x) \): \[ f(x) = (x + 5)\left(3x^2 - 10x + 25\right). \] Finally: B. There is only one rational zero, \( -5 \), and the other zeros are \( \frac{5 + 5i\sqrt{2}}{3}, \frac{5 - 5i\sqrt{2}}{3} \). Diving into a bit of historical context, polynomial equations have fascinated mathematicians for centuries, with roots traced back to ancient civilizations. The Babylonians had methods for solving quadratic equations, while Indian mathematicians expanded on these concepts, leading to the development of formulas we now consider standard. This journey through mathematical history not only showcases human ingenuity but also emphasizes the collaborative nature of knowledge throughout different cultures. For those who want to explore further, "The History of Mathematics: A Brief Course" by Roger L. Cooke offers a comprehensive overview of mathematical concepts through the ages. It highlights significant figures, theories, and the evolution of ideas, making it an exciting read for anyone looking to delve deeper into the fascinating world of mathematics and its development over time.