2
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expr=(2 (x1-x3) (x2^2-x3^2+y2^2-y3^2)+2 (x2-x3) (-x1^2+x3^2-y1^2+y3^2))/
  (4 (x3 (y1-y2)+x1 (y2-y3)+x2 (-y1+y3)));

Through[{Simplify,FullSimplify,Cancel}[expr]]

Obvious, this algebraic expression could be cancel common factors 2, I have tried Simplify, FullSimplify and Cancel, FullSimplify doesn't know simplify it, Cancel make it more complicated. enter image description here

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  • $\begingroup$ I usually use Factor in such situations. $\endgroup$ – Somos Jun 17 at 10:39
1
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Try

expr // Together

which eleminates Factor 2

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0
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This is not answer, but a workaround and an observation, as I do not know why Mathematica does not cancel the 2 out from the expression automatically.

But if you type this

ClearAll[a]
num = (a (x1 - x3) (x2^2 - x3^2 + y2^2 - y3^2) + a (x2 - x3) (-x1^2 + x3^2 - y1^2 + y3^2));
den = (2*a (x3 (y1 - y2) + x1 (y2 - y3) + x2 (-y1 + y3)));
num/den

You'll see it also does not cancel the a.

Mathematica graphics

So for a workaround, just do

Collect[num, a]/den

And now it is canceled giving

Mathematica graphics

Note, in my answer above, I assumed you wanted to keep the form of the numerator and denominator not expanded. If you do not care about this, then you could always expand and then simplify:

 Simplify[Expand[num]/Expand[den]]

Mathematica graphics

Which cancels the 2

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