Factorise :
x2+a2+1ax+1x^2 + \dfrac{a^2 + 1}{a}x + 1x2+aa2+1x+1
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Given,
=x2+a2+1ax+1=ax2+(a2+1)x+aa=ax2+a2x+x+aa=ax(x+a)+(x+a)a=(x+a)(ax+1)a=(x+a)ax+1a=(x+a)(x+1a).\phantom{=} x^2 + \dfrac{a^2 + 1}{a}x + 1 \\[1em] = \dfrac{ax^2 + (a^2 + 1)x + a}{a} \\[1em] = \dfrac{ax^2 + a^2x + x + a}{a} \\[1em] = \dfrac{ax(x + a) + (x + a)}{a} \\[1em] = \dfrac{(x + a)(ax + 1)}{a} \\[1em] = (x + a)\dfrac{ax + 1}{a} \\[1em] = (x + a)\Big(x + \dfrac{1}{a}\Big).=x2+aa2+1x+1=aax2+(a2+1)x+a=aax2+a2x+x+a=aax(x+a)+(x+a)=a(x+a)(ax+1)=(x+a)aax+1=(x+a)(x+a1).
Hence, x2+a2+1ax+1=(x+a)(x+1a).x^2 + \dfrac{a^2 + 1}{a}x + 1 = (x + a)\Big(x + \dfrac{1}{a}\Big).x2+aa2+1x+1=(x+a)(x+a1).
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