Factorise :
x2+1x2+2−5x−5xx^2 + \dfrac{1}{x^2} + 2 - 5x - \dfrac{5}{x}x2+x21+2−5x−x5
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x2+1x2+2−5x−5x=(x2+1x2+2)−(5x+5x)=(x2+1x2+2×x×1x)−5(x+1x)=(x+1x)2−5(x+1x)=(x+1x)[(x+1x)−5]=(x+1x)(x+1x−5)x^2 + \dfrac{1}{x^2} + 2 - 5x - \dfrac{5}{x}\\[1em] = \Big(x^2 + \dfrac{1}{x^2} + 2\Big) - \Big(5x + \dfrac{5}{x}\Big)\\[1em] = \Big(x^2 + \dfrac{1}{x^2} + 2 \times x \times \dfrac{1}{x}\Big) - 5\Big(x + \dfrac{1}{x}\Big)\\[1em] = \Big(x + \dfrac{1}{x}\Big)^2 - 5\Big(x + \dfrac{1}{x}\Big)\\[1em] = \Big(x + \dfrac{1}{x}\Big)\Big[\Big(x + \dfrac{1}{x}\Big) - 5\Big]\\[1em] = \Big(x + \dfrac{1}{x}\Big)\Big(x + \dfrac{1}{x} - 5\Big)x2+x21+2−5x−x5=(x2+x21+2)−(5x+x5)=(x2+x21+2×x×x1)−5(x+x1)=(x+x1)2−5(x+x1)=(x+x1)[(x+x1)−5]=(x+x1)(x+x1−5)
Hence, x2+1x2+2−5x−5x=(x+1x)(x+1x−5)x^2 + \dfrac{1}{x^2} + 2 - 5x - \dfrac{5}{x} = \Big(x + \dfrac{1}{x}\Big)\Big(x + \dfrac{1}{x} - 5\Big)x2+x21+2−5x−x5=(x+x1)(x+x1−5).
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