How do you divide $(x^2-1/6x-1/6)/(x-1/2)$ ?

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Answer 1 · 2016-04-16T09:39:58

$(x^2-1/6x-1/6)/(x-1/2)=x+1/3$

$(x^2-1/6x-1/6)/(x-1/2)=(1/6(6x^2-x-1))/(1/2(2x-1))=(2(6x^2-x-1))/(6(2x-1))$

= $(6x^2-x-1)/(3(2x-1)$

Now splitting the middle term $-x$ in numerator to $-3x+2x$ , we get

= $(6x^2-3x+2x-1)/(3(2x-1))$

= $(3x(2x-1)+1(2x-1))/(3(2x-1))$

= $((3x+1)(2x-1))/(3(2x-1))=(3x+1)/3=x+1/3$

How do you divide (x^2-1/6x-1/6)/(x-1/2)(x^2-1/6x-1/6)/(x-1/2)?