Not sure how this works! Apparently it can be done in 5-6 multiplications
Show how to multiply two degree 2 polynomials using fewer multiplications of coefficients than the naive algorithm.
I can't find the way using only 5-6, but here is a way to do it in $7$.
Suppose we are multiplying together $ax^2 + bx + c$ and $dx^2 + ex + f$. Compute $(a+b+c)(d+e+f) = ad + ae + af + bd + be + bf + cd + ce + cf$. This takes $1$ multiplication. Now compute $ad,ae,bd,bf,ce,cf$. This brings up up to $7$. The coefficients of the product are simply $ad, ae + bd, be + af + cd, bf + ce, cf$. By straightfoward additions we know $ad, ae + bd, bf + ce, cf$ already. This coefficient $be + af + cd$ can be computed as $(a+b+c)(d+e+f) - ad - ae - bd - bf - ce - cf$, thus we're done and we only need $7$ multiplications.