MHB Yes, your factorization is correct.

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The expression A^2 - B^2 + 16A + 64 can be factored using the difference of squares method. Initially, it was incorrectly grouped, but the correct approach reveals it can be rewritten as (A + 8)^2 - B^2. This leads to the final factorization of (A + 8 + B)(A + 8 - B). The discussion highlights the importance of recognizing perfect squares in factoring. More questions from the same textbook section are anticipated in future posts.
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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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