# Tag Info

## New answers tagged factorization

2

I think this is an exponent. The answer is: 2*(x+1)^2*(2+x)^2*(3+x)

1

Another way assuming $r\ge 0,s\ge 0,t\ge 0$. expr = 1 - 2 r + r^2 - 2 s + 2 r s + s^2 - 2 t + 2 r t + 2 s t - 4 r s t + t^2; EXPR = expr /. {s -> S^2 , r -> R^2, t -> T^2} // Factor EXPR /.{S -> Sqrt[s], R -> Sqrt[r], T -> Sqrt[t]}

1

Perhaps this is what you can use in this case ex = 1-2r+r^2-2s+2r s+s^2-2t+2r t+2s t-4r s t+t^2; (ex/.r s t->u^2//Factor)/.u->Sqrt[r s t] which returns (-1 + r + s + t - 2*Sqrt[r*s*t])*(-1 + r + s + t + 2*Sqrt[r*s*t] I don't think this can be generalized. Any expression with more than one term can be written as  a-b = (\sqrt{a})^2 - (\sqrt{b})^2= (\...

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