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Listen Factor \( 3 p^{5}-192 p^{3} \) completely. \( 3 p^{5}-192 p^{3}=3 p^{3}\left(p^{2}-64\right) \)

Ask by French Park. in the United States
Feb 21,2025

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\( 3p^{5} - 192p^{3} = 3p^{3}(p - 8)(p + 8) \)

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To factor \( 3 p^{5}-192 p^{3} \) completely, we can start by recognizing that both terms share a common factor of \( 3p^{3} \). Factoring this out gives us: \[ 3 p^{5}-192 p^{3} = 3 p^{3}(p^{2}-64). \] Now, we notice that \( p^{2}-64 \) is a difference of squares, which can be further factored as: \[ p^{2}-64 = (p-8)(p+8). \] Putting it all together, we have: \[ 3 p^{5}-192 p^{3} = 3 p^{3}(p-8)(p+8). \] How neat is that? You've transformed a polynomial into a beautifully factored form! If we take a quick step back in time, the development of algebra allowed for such factorizations to be understood systematically. The concept of factoring polynomials can be traced back to ancient mathematicians, who devised methods to simplify expressions and solve equations. Muhammad ibn Musa al-Khwarizmi's work in the 9th century laid the groundwork for many algebraic principles still in use today. Fascinating, isn't it? This historical context showcases the evolution of mathematical thought, leading us to the efficient techniques we use to factor polynomials today.

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