Pregunta
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1. \( 25 a^{2}-16 \) 3. \( 2 a^{2}+2 b^{2} \) 5. \( x^{2} y^{2}-1 \) 7. \( 28 x^{3}-63 x b^{2} \) 9. \( (2 x+3 y)^{2}-9 \) 11. \( 4 x^{2}-(3 x-2 y)^{2} \) 13. \( 25 a^{2}(a-3 b)+9(3 b-a) \)

Ask by Hamilton Lyons. in South Africa
Feb 03,2025

Solución de tutoría real

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Here are the factored forms of the given expressions: 1. \(25a^{2} - 16 = (5a - 4)(5a + 4)\) 2. \(2a^{2} + 2b^{2} = 2(a^{2} + b^{2})\) 3. \(x^{2}y^{2} - 1 = (xy - 1)(xy + 1)\) 4. \(28x^{3} - 63xb^{2} = 7x(2x - 3b)(2x + 3b)\) 5. \((2x + 3y)^{2} - 9 = (2x + 3y - 3)(2x + 3y + 3)\) 6. \(4x^{2} - (3x - 2y)^{2} = (-x + 2y)(5x - 2y)\) 7. \(25a^{2}(a - 3b) + 9(3b - a) = (a - 3b)(5a - 3)(5a + 3)\) If you need further explanation on any of these, feel free to ask!

Solución

Certainly! Let's factor each of the given expressions step by step. --- **1. \( 25a^{2} - 16 \)** This is a **difference of squares**: \[ 25a^{2} - 16 = (5a)^{2} - 4^{2} = (5a - 4)(5a + 4) \] --- **3. \( 2a^{2} + 2b^{2} \)** Factor out the common term **2**: \[ 2a^{2} + 2b^{2} = 2(a^{2} + b^{2}) \] *Note:* \(a^{2} + b^{2}\) cannot be factored further over the real numbers. --- **5. \( x^{2}y^{2} - 1 \)** This is another **difference of squares**: \[ x^{2}y^{2} - 1 = (xy)^{2} - 1^{2} = (xy - 1)(xy + 1) \] --- **7. \( 28x^{3} - 63xb^{2} \)** First, factor out the greatest common factor **7x**: \[ 28x^{3} - 63xb^{2} = 7x(4x^{2} - 9b^{2}) \] Notice that \(4x^{2} - 9b^{2}\) is a **difference of squares**: \[ 4x^{2} - 9b^{2} = (2x)^{2} - (3b)^{2} = (2x - 3b)(2x + 3b) \] So, the fully factored form is: \[ 28x^{3} - 63xb^{2} = 7x(2x - 3b)(2x + 3b) \] --- **9. \( (2x + 3y)^{2} - 9 \)** This is a **difference of squares**: \[ (2x + 3y)^{2} - 3^{2} = (2x + 3y - 3)(2x + 3y + 3) \] --- **11. \( 4x^{2} - (3x - 2y)^{2} \)** Again, a **difference of squares**: \[ 4x^{2} - (3x - 2y)^{2} = (2x)^{2} - (3x - 2y)^{2} = [2x - (3x - 2y)][2x + (3x - 2y)] \] Simplify each binomial: \[ [2x - 3x + 2y][2x + 3x - 2y] = (-x + 2y)(5x - 2y) \] *Alternatively written as:* \[ (2y - x)(5x - 2y) \] --- **13. \( 25a^{2}(a - 3b) + 9(3b - a) \)** First, notice that \(9(3b - a) = -9(a - 3b)\). Factor out \((a - 3b)\): \[ 25a^{2}(a - 3b) - 9(a - 3b) = (a - 3b)(25a^{2} - 9) \] Now, factor \(25a^{2} - 9\) as a **difference of squares**: \[ 25a^{2} - 9 = (5a)^{2} - 3^{2} = (5a - 3)(5a + 3) \] So, the fully factored form is: \[ 25a^{2}(a - 3b) + 9(3b - a) = (a - 3b)(5a - 3)(5a + 3) \] --- **Summary of Factored Forms:** 1. \(25a^{2} - 16 = (5a - 4)(5a + 4)\) 3. \(2a^{2} + 2b^{2} = 2(a^{2} + b^{2})\) 5. \(x^{2}y^{2} - 1 = (xy - 1)(xy + 1)\) 7. \(28x^{3} - 63xb^{2} = 7x(2x - 3b)(2x + 3b)\) 9. \((2x + 3y)^{2} - 9 = (2x + 3y - 3)(2x + 3y + 3)\) 11. \(4x^{2} - (3x - 2y)^{2} = (-x + 2y)(5x - 2y)\) 13. \(25a^{2}(a - 3b) + 9(3b - a) = (a - 3b)(5a - 3)(5a + 3)\) --- If you have any specific questions about these factorizations or need further assistance, feel free to ask!

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Extra Insights

The expression \( 25 a^{2}-16 \) is a classic example of a difference of squares, which can be factored as \( (5a - 4)(5a + 4) \). This is a useful technique in algebra, simplifying quadratic equations and making them easier to solve or analyze. For \( 2 a^{2} + 2 b^{2} \), one might mistakenly look for a common factor only. However, this expression doesn't factor into two binomials over the reals because it represents a sum of squares. Instead, you can factor out the 2, giving \( 2(a^{2} + b^{2}) \). Just remember, there’s no easy way to factor sums of squares using real numbers!

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