We’re learning a lot of algorithm in my algebre et calcul formel class. One of them is the Toom-Cook algorithm used for multiplication of large integers.

I found a super simple explanation of it on a forum, it helps:

Say, we want to multiply 23 times 35.
We write,
p(x) = 2x + 3,
q(x) = 3x + 5.
We are using our realization that any integer can be written as a polynomial.
Here, p(x), represents 23, and q(x), represents 35, when x equals 10.
We write,
p(x)q(x) = r(x).
That is, p(x) times q(x), equals r(x).
So,
(2x + 3)(3x + 5) = ax^2 + bx + c = r(x).
Now,
p(0)q(0) = r(0).
So,
(20 + 3)(30 + 5) = a0 + b0 + c.
Therefore,
c = 15.
Now,
p(1)q(1) = r(1).
Therefore, when we do the substitutions (for x and c),
a + b = 25.
Now,
p(-1)q(-1) = r(-1).
Therefore, when we do the substitutions (for x and c),
a - b = -13.
Now, we already know c, and we just need to find a and b.
We have two linear equations and two unknowns,
a + b = *25,
a - b = -13.
We just add the two equations and we get,
2a = 12.
Therefore,
a = 6.
Now, we can substitute 6 for a in,
a + b = 25,
and we get,
b = 19.
So,
r(x) = 6x^2 + 19x + 15.
Now, we substitute 10 for x in r(x), and we are done,
r(10) = 600 + 190 + 15 = 805.
Believe it or not!