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). This solution is derived by first grouping the terms into Group A (a^2t^2 + b^2t^2) and Group B (-cb^2 - ca^2), leading to the factorization of t^2(a^2 + b^2) - c(a^2 + b^2). The correctness of this factorization is confirmed in the discussion.

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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.
 

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