Yes, your factorization is correct.

[mathdad]

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?

 

[kaliprasad]

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

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

 

[mathdad]

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?

 

[HallsofIvy]

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"

 

[mathdad]

More factoring questions will be posted tomorrow from section 1.3 in David Cohen's precalculus textbook, third edition.