Next we examine a polynomial of degree 3:
poly: x^3 + x^2 - 8*x - 12;
The computation of the zeroes is straightforward:
sol: solve (poly=0, x);
[x = 3, x = - 2]
The answer is again written as a list of two equations. We know that for an equation of third degree we can expect three solutions, so we assume that one of the solutions is a double zero. Which one?
To answer that question, we type in multiplicities and obtain this list:
This means that the first equation in the solution list has multiplicity one, while the second solution has multiplicity two.
To proceed, we rewrite the equations as terms:
terms: map( lambda( [eq], lhs(eq) - rhs(eq)), sol);
[x - 3, x + 2]
Now we have to use the multiplicities to form the correct factors:
map(lambda ( [term, exponent], term^exponent), terms, m);
2 [x - 3, (x + 2) ]
This time we used map with a symbolic function with two arguments. For such a function, we have also to provide two lists of arguments.
No we can again apply multiplication:
2 (x - 3) (x + 2)
This is a polynomial in factored form. Expansion gives us the initial polynomial:
3 2 x + x - 8 x - 12
It is of course possible to put all these computations into one single expression:
expand(apply("*", map( lambda( [eq, exponent], (lhs(eq) - rhs(eq))^exponent), sol, multiplicities ) ) );
The innermost expression (here the map) is evaluated first. This is not a surprise: The result of map is needed to perform the apply and the result of apply is needed to perform the expand.