Factorise :
x2+1x2−2−3x+3xx^2 + \dfrac{1}{x^2} - 2 - 3x + \dfrac{3}{x}x2+x21−2−3x+x3.
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x2+1x2−2−3x+3x=x2+1x2−2×x×1x−3x+3x=(x−1x)2−3(x−1x)=(x−1x)[(x−1x)−3]=(x−1x)(x−1x−3)x^2 + \dfrac{1}{x^2} - 2 - 3x + \dfrac{3}{x}\\[1em] = x^2 + \dfrac{1}{x^2} - 2 \times x \times \dfrac{1}{x} - 3x + \dfrac{3}{x}\\[1em] = \Big(x - \dfrac{1}{x}\Big)^2 - 3\Big(x - \dfrac{1}{x}\Big)\\[1em] = \Big(x - \dfrac{1}{x}\Big)\Big[\Big(x - \dfrac{1}{x}\Big) - 3\Big]\\[1em] = \Big(x - \dfrac{1}{x}\Big)\Big(x - \dfrac{1}{x} - 3\Big)x2+x21−2−3x+x3=x2+x21−2×x×x1−3x+x3=(x−x1)2−3(x−x1)=(x−x1)[(x−x1)−3]=(x−x1)(x−x1−3)
Hence, x2+1x2−2−3x+3x=(x−1x)(x−1x−3)x^2 + \dfrac{1}{x^2} - 2 - 3x + \dfrac{3}{x} = \Big(x - \dfrac{1}{x}\Big)\Big(x - \dfrac{1}{x} - 3\Big)x2+x21−2−3x+x3=(x−x1)(x−x1−3).
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If x - y + z = 5 and x2 + y2 + z2 = 49, find the value of : zx - xy - yz.
x(a - 5) + y(5 - a).
x2−2x−9x^2 - 2x - 9x2−2x−9.
13x2−8x\dfrac{1}{3}x^2 - \dfrac{8}{x}31x2−x8
