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.
 
Thread 'Erroneously finding discrepancy in transpose rule'

Thread 'Erroneously finding discrepancy in transpose rule'

Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...
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