Hi René
it's just the variable, in your case \(x\). You can find all of this in the docs or on GitHub.
Stephan
Hello Stephan
Thanks for your help and the link to the documentation.
As far as I can tell, the factorisation now works in principle.
Thanks for your help and the link to the documentation.
As far as I can tell, the factorisation now works in principle.
However, \( 2\cdot x^2-2 \) (`ex1`) is corrected as wrong when written as \( (-2)\cdot(-x+1)\cdot(x+1) \) (`ans1`). Is it somehow possible
i) to recognise whether `ans1` is a product and
ii) `ans1` consists of at least `factor(ex1)` factors?
i) to recognise whether `ans1` is a product and
ii) `ans1` consists of at least `factor(ex1)` factors?
This would then correct answers like \( (-2)\cdot(-x+1)\cdot(x+1) \) or \( (-1)\cdot2\cdot(1-x)\cdot(x+1) \) as correct.
Possibly with something like `length(part(factor(ex1)));`.
Hi,
FacForm should accept \( (-2)\cdot (-x+1)\cdot (x+1)\), EqualComAss will not.
Of course you could use something like
Depending on what you want, it's probably easier and more robust to use EqualComAss to first check on \( 2\cdot (x-1)\cdot (x+1)\) and if false also on \( (-2)\cdot (-x+1)\cdot (x+1)\).
Stephan
FacForm should accept \( (-2)\cdot (-x+1)\cdot (x+1)\), EqualComAss will not.
Of course you could use something like
is (safe_op(ans1)="*") to determine if ans1 is a product and length(args(ans1)) to count the factors. But you would have to deal with many different cases if you want to implement this in your feedback variables (\(2\cdot \left(\frac 1 2\right) \cdot (2 x^2 - 2)\) is a product with 3 factors, too.). Depending on what you want, it's probably easier and more robust to use EqualComAss to first check on \( 2\cdot (x-1)\cdot (x+1)\) and if false also on \( (-2)\cdot (-x+1)\cdot (x+1)\).
Stephan