MHB How do you factor expressions like 3(x + h)^4 - 48(x + h)^2?

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The expression 3(x + h)^4 - 48(x + h)^2 can be factored by first extracting 3(x + h)^2, resulting in 3(x + h)^2[(x + h)^2 - 16]. This further simplifies to 3(x + h)^2[(x + h) - 4][(x + h) + 4]. The discussion highlights the complexity of factoring problems found in Precalculus by David Cohen, particularly in Section 1.3, which includes many challenging examples. Participants express a commitment to sharing additional tricky factoring problems in future posts. The focus remains on mastering these advanced factoring techniques.
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Precalculus by David Cohen, 3rd Edition
Chapter 1, Section 1.3.
Question 32c.

Factor the expression.

3(x + h)^4 - 48(x + h)^2

Solution:

Factor out 3(x + h)^2.

3(x + h)^2[(x + h)^2 - 16]

Simplify the quantity in the brackets.

3(x + h)^2[(x + h) - 4][(x - h) + 4]

Is this right?
 
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RTCNTC said:
Precalculus by David Cohen, 3rd Edition
Chapter 1, Section 1.3.
Question 32c.

Factor the expression.

3(x + h)^4 - 48(x + h)^2

Solution:

Factor out 3(x + h)^2.

3(x + h)^2[(x + h)^2 - 16]

Simplify the quantity in the brackets.

3(x + h)^2[(x + h) - 4][(x - h) + 4]

Is this right?

right
 
Section 1.3 has what appears to be endless factoring questions. I will post many factoring problems in the coming days. I am talking about "tricky" factoring problems not factor 3a + a.
 
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