Yes, your factorization is correct.

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Discussion Overview

The discussion revolves around the factorization of the expression A^2 - B^2 + 16A + 64, as presented in a precalculus textbook. Participants explore different methods of factoring, specifically using the grouping method and the concept of perfect squares.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes a factorization approach using grouping, suggesting that A^2 - B^2 can be factored as (A - B)(A + B) and 16A + 64 as 16(A + 4).
  • Another participant challenges this approach, asserting that the expression is not factored correctly and instead presents an alternative factorization as (A + 8 + B)(A + 8 - B) after rewriting the expression as a difference of squares.
  • A question is raised regarding the placement of B^2 and the arrangement of terms in the expression, indicating a need for clarity in the factorization process.
  • A later reply notes the convenience of recognizing (A + 8)^2 as a perfect square, suggesting that this recognition aids in forming a difference of squares.

Areas of Agreement / Disagreement

Participants express disagreement regarding the correctness of the initial factorization approach. Multiple competing views on how to factor the expression remain unresolved.

Contextual Notes

Participants do not reach a consensus on the correct method of factorization, and there are unresolved questions about the arrangement of terms in the expression.

Who May Find This Useful

Students studying precalculus, particularly those interested in factoring techniques and the properties of polynomials.

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Precalculus by David Cohen, 3rd Edition
Chapter 1, Section 1.3.
Question 50.

Factor the expression.

A^2 - B^2 + 16A + 64

Factor by grouping method.

Group A = A^2 - B^2

Group A = (A - B)(A + B)

Group B = 16A + 64

Group B = 16(A + 4)

Group A + Group B

(A - B)(A + B) + 16(A + 4)

Correct?
 
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RTCNTC said:
Precalculus by David Cohen, 3rd Edition
Chapter 1, Section 1.3.
Question 50.

Factor the expression.

A^2 - B^2 + 16A + 64

Factor by grouping method.

Group A = A^2 - B^2

Group A = (A - B)(A + B)

Group B = 16A + 64

Group B = 16(A + 4)

Group A + Group B

(A - B)(A + B) + 16(A + 4)

Correct?

No.
above is not factors. it is sum of 2 expressions

$A^2 - B^2 + 16A + 64 = A^2 + 16A + 64 - B^2 = (A+8)^2 - B^2 = (A+8+B)(A+8-B)$
 
Last edited:
kaliprasad said:
No.
above is not factors. it is sum of 2 expressions

$A^2 - B^2 + 16A + 64 = A^2 + 16A + 64 - B^2 = (A+8)^2 - B^2 = (A+8+B)(A+8-B)$

Why did you put B^2 to the far right?

Why did you put 16A + 64 in the center between A^2 and B^2?
 
Because it was convenient. Kaliprasad recognized that A^2a+ 16A+ 64= (A+ 8)^2 is itself a "perfect square" so this could be written as a single "difference of squares"
 
More factoring questions will be posted tomorrow from section 1.3 in David Cohen's precalculus textbook, third edition.
 

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