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Simplify the following expressions fully: \( \begin{array}{ll}1.1 .1 & 3 x^{3}-2 x^{2}(x+5) \\ 1.1 .2 & (2 x-1)(2 x+1) \\ 1.1 .3 & \frac{1}{4} x\left(4 x^{-1}-8 x\right) \\ 1.2 & \text { Factorise the following expression fully: } \\ y^{2}(y-2)+x^{2}(2-y)\end{array} \)

Ask by Savage Beck. in South Africa
Mar 10,2025

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1.1.1 \( x^3 - 10x^2 = x^2(x - 10) \) 1.1.2 \( (2x - 1)(2x + 1) = 4x^2 - 1 \) 1.1.3 \( \frac{1}{4}x(4x^{-1} - 8x) = 1 - 2x^2 \) 1.2 \( y^2(y - 2) + x^2(2 - y) = (y - 2)(y - x)(y + x) \)

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Beyond the Answer

To simplify the given expressions, let's break them down one by one: 1. **For \( 3x^3 - 2x^2(x + 5) \):** First, distribute the \( -2x^2 \) through \( (x + 5) \): \[ -2x^2 \cdot x - 2x^2 \cdot 5 = -2x^3 - 10x^2. \] Now combine like terms: \[ 3x^3 - 2x^3 - 10x^2 = x^3 - 10x^2. \] So, the simplified expression is: \[ x^3 - 10x^2. \] 2. **For \( (2x - 1)(2x + 1) \):** This expression is a difference of squares, and it simplifies as follows: \[ (2x)^2 - (1)^2 = 4x^2 - 1. \] So, the simplified expression is: \[ 4x^2 - 1. \] 3. **For \( \frac{1}{4}x(4x^{-1} - 8x) \):** Distributing \( \frac{1}{4}x \): \[ \frac{1}{4}x \cdot 4x^{-1} - \frac{1}{4}x \cdot 8x = x - 2x^2. \] So, the simplified expression is: \[ x - 2x^2. \] 4. **For \( y^2(y - 2) + x^2(2 - y) \):** Distributing \( y^2 \) and \( x^2 \): \[ y^3 - 2y^2 + 2x^2 - xy^2. \] Rearranging gives: \[ y^3 - 2y^2 - xy^2 + 2x^2. \] The simplified expression could be re-organized as: \[ y^2(y - 2 - x) + 2x^2. \] In summary, the fully simplified forms are: 1. \( x^3 - 10x^2 \) 2. \( 4x^2 - 1 \) 3. \( x - 2x^2 \) 4. \( y^2(y - 2 - x) + 2x^2 \)

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