1.2. Factorise fully: 1.2.1. \( a^{2}-2 a-a x+2 x \) 1.2.2. \( -x y-(y-x) b+b^{2} \) 1.2.3. \( 6 p+40-p^{2} \) 1.2.4. \( a^{2}-2 a b+b^{2}-100 c^{2} \) 1.2.5. \( 2 x^{2}-x-6 \)
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Extra Insights
To factorise the given expressions fully: 1.2.1. \( a^{2}-2a-ax+2x \) Rearranging gives us \( a^{2} - 2a - ax + 2x = a(a-2-x) + 2x \). We can rewrite it as \( a(a-2-x) + 2(x)\). Thus, the expression becomes \( (a-2)(a-x+2) \). 1.2.2. \( -xy-(y-x)b+b^{2} \) Rearranging gives us \( -xy - by + bx + b^{2} \) or \( -y(x-b) + b(x+b) \). Factoring out common terms leads to \( -(x-b)(y+b) \). 1.2.3. \( 6p + 40 - p^{2} \) Rearranging gives us \( -p^{2} + 6p + 40 = -(p^{2} - 6p - 40) \). Factoring the quadractic gives us \( -(p-10)(p+4) \). Thus, the expression becomes \( -(p-10)(p+4) \). 1.2.4. \( a^{2}-2ab+b^{2}-100c^{2} \) We'll notice the first part is a perfect square: \( (a-b)^{2} - (10c)^{2} \), which is a difference of squares. It can be factored as \( ((a-b)-10c)((a-b)+10c) \). 1.2.5. \( 2x^{2}-x-6 \) We will factor the quadratic. Start with \( (2x+3)(x-2) \) as it satisfies \( 2x^2 - x - 6 \). Thus, the expression is \( 2(x + 3)(x - 2) \).
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