I am trying to solve this problem here but have to understand basics. Suppose I have a multiplication such as $a*(b+c+d+...+x)$. How can I multiply it to the result $a*b+a*c+...+a*x$?


[Not working] Reduce[(a + b)*c]

[Not working] Simplify[(a + b)*c] or FullSimplify[(a + b)*c]

  • 1
    $\begingroup$ Try Distribute[a * (x + y + z)]. $\endgroup$ Dec 9, 2012 at 12:51
  • $\begingroup$ @b.gatessucks plesae, make it into an answer. It worked? It is giving wrong result with "Distribute[(a + b)*(a + b)]" , why? $\endgroup$
    – hhh
    Dec 9, 2012 at 12:52
  • 4
    $\begingroup$ Or Expand[a*(b + c)]. $\endgroup$ Dec 9, 2012 at 12:56
  • 1
    $\begingroup$ Your new example is explicitly covered in the documentation. You should read it. $\endgroup$ Dec 9, 2012 at 13:08
  • 2
    $\begingroup$ There are many ways besides Expand, e.g. PolynomialReduce[a (b + c), 1, c][[1, 1]]. Take a look here : mathematica.stackexchange.com/questions/9111/… $\endgroup$
    – Artes
    Dec 9, 2012 at 13:35

2 Answers 2


I know this has been answered in the comments, but just for the sake of having an actual answer..




While covered by the comments and another answer, I would like to expand on this a bit. The word "simplify" has very different connotations between what you intend and what Mathematica intends. For Mathematica, it means simply the least complex solution by some measure. This defaults to the number of subexpressions and integer digits, but it can be set to whatever you wish in an attempt to coerce Simplify to generate a specific form (or avoid others), such as this example from the docs

f[e_] := 100 Count[e, _ChebyshevT, {0, Infinity}] + LeafCount[e]
FullSimplify[ChebyshevT[n, x], ComplexityFunction -> f]
(* Cos[n ArcCos[x]] *)

where ChebyshevT is being made more expensive than its alternative forms. I would also look at the example for Abs below that. Using this definition, then, the suggestions of Expand make sense as it is not simplest (least complex) form.

Your use of the word simplify is very different. In assigning problems to students, we use the word simplify imprecisely, and what we mean is transform it into a specific form which is not necessarily the simplest in form. I think this is where the confusion lies; this is not what Mathematica means, but it may be convinced that it is correct.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.