# Multiplying Polynomials with fewer coefficient multiplications

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.

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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.
The passage describes a method for multiplying two degree 2 polynomials using fewer coefficient multiplications than the naive algorithm. It provides a specific method that uses 7 multiplications and explains the steps involved.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Multiplying Polynomials with fewer coefficient multiplications

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.
user
The passage describes a method for multiplying two degree 2 polynomials using fewer coefficient multiplications than the naive algorithm. It provides a specific method that uses 7 multiplications and explains the steps involved.

TYPE: SINGLE_QUESTION_WITH_ANSWER