text: # 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.
explanation: The document discusses a method for multiplying two degree 2 polynomials using fewer coefficient multiplications than the naive algorithm. It starts with a question about how to achieve this with 5-6 multiplications and then provides a detailed explanation of a method that uses 7 multiplications. The document contains a question and an answer.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # 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.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document discusses a method for multiplying two degree 2 polynomials using fewer coefficient multiplications than the naive algorithm. It starts with a question about how to achieve this with 5-6 multiplications and then provides a detailed explanation of a method that uses 7 multiplications. The document contains a question and an answer.
  
  Label: question-and-answer