How do you factor the trinomial $x^{2} + 1 - x$ $x^2 +1 - x$ ?

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Answer 1 · 2017-08-17T11:03:55

$(x - \frac{1}{2} + i \frac{\sqrt{3}}{2}) (x - \frac{1}{2} - i \frac{\sqrt{3}}{2}) .$ $(x-1/2+isqrt3/2)(x-1/2-isqrt3/2).$

$x^{2} + 1 - x = x^{2} - x + 1 .$ $x^2+1-x=x^2-x+1.$

Since, $x^{2} - x = x^{2} - 2 (\frac{1}{2}) x,$ $x^2-x=x^2-2(1/2)x,$ to make it a perfect square, we need the

last term ${(\frac{1}{2})}^{2} = \frac{1}{4} .$ $(1/2)^2=1/4.$

Hence, $x^{2} - x + 1 = x^{2} - x + \frac{1}{4} + \frac{3}{4},$ $x^2-x+1=x^2-x+1/4+3/4,$

$= {(x - \frac{1}{2})}^{2} - (- \frac{3}{4}),$ $=(x-1/2)^2-(-3/4),$

$= {(x - \frac{1}{2})}^{2} - ⎧ ⎨ ⎩ i^{2} \cdot {(\frac{\sqrt{3}}{2})}^{2} ⎫ ⎬ ⎭,$ $=(x-1/2)^2-{i^2*(sqrt3/2)^2},$

$= {(x - \frac{1}{2})}^{2} - {(i \frac{\sqrt{3}}{2})}^{2},$ $=(x-1/2)^2-(isqrt3/2)^2,$

$= (x - \frac{1}{2} + i \frac{\sqrt{3}}{2}) (x - \frac{1}{2} - i \frac{\sqrt{3}}{2}),$ $=(x-1/2+isqrt3/2)(x-1/2-isqrt3/2),$ is the desired

factorisation.

How do you factor the trinomial x^2 +1 - xx2+1−xx^2 +1 - x?