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1. \( \frac{x^{2}}{16}-\frac{1}{25} \) 2. \( 15 x^{2}-36 x+12 \) 3. \( 2 x^{2}-7 x-4 \) 4. \( 3 x^{2}-8 x+4 \) 5. \( 6 x^{2}+4 x-2 \) 6. \( 2 x^{2}+3 x-5 \) 7. \( 9 x^{2}+27 x+8 \) 8. \( 12 x^{2}+4 x-5 \) 9. \( 24 x^{2}+x-3 \)

Ask by Byrd Morgan. in South Africa
Mar 16,2025

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Answer

Here are the factorizations for the given expressions: 1. \( \frac{x^{2}}{16}-\frac{1}{25} \) – No factorization found. 2. \( 15 x^{2}-36 x+12 \) – \( 3(x-2)(5x-2) \) 3. \( 2 x^{2}-7 x-4 \) – \( (x-4)(2x+1) \) 4. \( 3 x^{2}-8 x+4 \) – \( (x-2)(3x-2) \) 5. \( 6 x^{2}+4 x-2 \) – \( 2(x+1)(3x-1) \) 6. \( 2 x^{2}+3 x-5 \) – \( (x-1)(2x+5) \) 7. \( 9 x^{2}+27 x+8 \) – \( (3x+1)(3x+8) \) 8. \( 12 x^{2}+4 x-5 \) – \( (2x-1)(6x+5) \) 9. \( 24 x^{2}+x-3 \) – \( (3x-1)(8x+3) \)

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Extra Insights

For the first expression, \( \frac{x^{2}}{16}-\frac{1}{25} \), you may notice that it's structured like the difference of two squares. This means it can be factored or rearranged to find roots more easily. By rewriting it in a more recognizable format, you can derive values of \( x \) where this expression equals zero, which can help in real-world scenarios, like optimizing area in geometry. Moving on to quadratic expressions such as \( 15x^{2}-36x+12 \), they often represent parabolas in coordinate geometry. In practical terms, these equations can model various phenomena, from projectile motion to profit maximization in economics. Being able to factor these or apply the quadratic formula lets you uncover important insights, such as the vertex (which provides maximal or minimal value) or roots (when the graph intersects the x-axis).

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