text: ## Algebra 1
  
  $y^{2}$-16y+64 Using the rule of $a^{2}$-2ab+$b^{2}$= $(a-b)^{2}$ In this case, the a= y and b= 8 as substituting these gives us the original polynomial $a^{2}$-2ab+$b^{2}$ $y^{2}$-2(y)(8)+$8^{2}$ $y^{2}$-16y+64 Thus: $y^{2}$-16y+64 = $(y-8)^{2}$ = (y-8)(y-8)
explanation: The document provides a step-by-step solution to factor the polynomial \( y^2 - 16y + 64 \) using the rule \( a^2 - 2ab + b^2 = (a - b)^2 \). It shows the substitution of \( a = y \) and \( b = 8 \) and concludes with the factored form \( (y - 8)^2 \).
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: ## Algebra 1
  
  $y^{2}$-16y+64 Using the rule of $a^{2}$-2ab+$b^{2}$= $(a-b)^{2}$ In this case, the a= y and b= 8 as substituting these gives us the original polynomial $a^{2}$-2ab+$b^{2}$ $y^{2}$-2(y)(8)+$8^{2}$ $y^{2}$-16y+64 Thus: $y^{2}$-16y+64 = $(y-8)^{2}$ = (y-8)(y-8)
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document provides a step-by-step solution to factor the polynomial \( y^2 - 16y + 64 \) using the rule \( a^2 - 2ab + b^2 = (a - b)^2 \). It shows the substitution of \( a = y \) and \( b = 8 \) and concludes with the factored form \( (y - 8)^2 \).
  
  Label: no-question