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Mathematics

Factorise :

a2+1a2−2−3a+3aa^2 + \dfrac{1}{a^2} - 2 - 3a + \dfrac{3}{a}

Factorisation

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Answer

Given,

=a2+1a2−2−3a+3a=(a−1a)2−3(a−1a)=(a−1a)(a−1a−3).\phantom{=}a^2 + \dfrac{1}{a^2} - 2 - 3a + \dfrac{3}{a} \\[1em] = \Big(a - \dfrac{1}{a}\Big)^2 - 3\Big(a - \dfrac{1}{a}\Big) \\[1em] = \Big(a - \dfrac{1}{a}\Big)\Big(a - \dfrac{1}{a} - 3\Big).

Hence, a2+1a2−2−3a+3a=(a−1a)(a−1a−3).a^2 + \dfrac{1}{a^2} - 2 - 3a + \dfrac{3}{a} = \Big(a - \dfrac{1}{a}\Big)\Big(a - \dfrac{1}{a} - 3\Big).

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