MHB Is there more factoring to come?

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The expression a^2t^2 + b^2t^2 - cb^2 - ca^2 can be factored by grouping into (t^2 - c)(a^2 + b^2). The first group, t^2(a^2 + b^2), is derived from a^2t^2 + b^2t^2, while the second group, -c(a^2 + b^2), comes from -cb^2 - ca^2. The solution confirms the correctness of the factoring process. Additional factoring topics are expected to be discussed in the coming days.
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Precalculus by David Cohen, 3rd Edition
Chapter 1, Section 1.3.
Question 48.

Factor the expression by grouping.

a^2t^2 + b^2t^2 - cb^2 - ca^2

Group A = a^2t^2 + b^2t^2

Group A = t^2(a^2 + b^2)

Group B = -cb^2 - ca^2

Group B = -c(b^2 + a^2)

Group B = -c(a^2 + b^2)

Group A - Group B

t^2(a^2 + b^2) - c(a^2 + b^2)

(t^2 - c)(a^2 + b^2)

Correct?
 
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RTCNTC said:
Precalculus by David Cohen, 3rd Edition
Chapter 1, Section 1.3.
Question 48.

Factor the expression by grouping.

a^2t^2 + b^2t^2 - cb^2 - ca^2

Group A = a^2t^2 + b^2t^2

Group A = t^2(a^2 + b^2)

Group B = -cb^2 - ca^2

Group B = -c(b^2 + a^2)

Group B = -c(a^2 + b^2)

Group A - Group B

t^2(a^2 + b^2) - c(a^2 + b^2)

(t^2 - c)(a^2 + b^2)

Correct?

yes
 
More factoring coming in the next few days.
 
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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