Question
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3) \( \left(\frac{1}{6}-\frac{1}{4}\right) \div\left(\frac{1}{3}+\frac{1}{4}\right) \) 4) \( 4 m^{2}+4 m+1 \) 5) \( (2 x+1)^{3} \) 6) \( \frac{1}{100}-p^{2} \) 7) \( 3 x y-6 x+5 m y-10 m \) 8) \( x^{2}-27+50 \) 9) \( m^{3}-1 \) \( (x+y)(x-y) \)

Ask by Ramos Gardner. in Colombia
Feb 03,2025

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3) \(-\frac{1}{7}\) 4) \( (2m + 1)^2 \) 5) \( 8x^3 + 12x^2 + 6x + 1 \) 6) \( \left(\frac{1}{10} - p\right)\left(\frac{1}{10} + p\right) \) 7) \( (3x + 5m)(y - 2) \) 8) \( x^2 + 23 \) 9) \( (m - 1)(m^2 + m + 1) \) \( (x + y)(x - y) = x^2 - y^2 \)

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The Deep Dive

Let’s break these down into fun and digestible insights! Start with the fraction calculations! When you’re working with fractions, it’s super helpful to find a common denominator. For example, in \( \left(\frac{1}{6}-\frac{1}{4}\right) \), the common denominator would be 12. This means you'd convert both fractions: \( \frac{1}{6} = \frac{2}{12} \) and \( \frac{1}{4} = \frac{3}{12} \). Now it’s easy to compute! Polynomials like \( 4 m^{2}+4 m+1 \) can be factored! The goal is to express it as a product of two binomials. Here, it factors neatly to \( (2m + 1)(2m + 1) \) or \( (2m + 1)^2 \). This often makes it simpler to analyze or solve, so keep an eye out for those patterns!

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