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Factorize completely:
(a) 3p² + 6p
(b) 4q² - 1

User Joshua Son
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1 Answer

1 vote

Final answer:

For expression (a), 3p² + 6p is factorized as 3p(p + 2). For expression (b), 4q² - 1 is a difference of squares and factorizes to (2q - 1)(2q + 1).

Step-by-step explanation:

The student has asked to factorize completely the following expressions: (a) 3p² + 6p and (b) 4q² - 1.

Factorization of 3p² + 6p

First, we identify a common factor in both terms of the expression. The common factor is 3p:

3p² + 6p = 3p(p + 2)

This expression cannot be factored further, so 3p(p + 2) is the completely factorized form of 3p² + 6p.

Factorization of 4q² - 1

The expression 4q² - 1 is a difference of squares and can be factored into:

4q² - 1 = (2q - 1)(2q + 1)

The factorization is now complete.

In answering similar questions, we should eliminate terms wherever possible to simplify the algebra and check the answer to see if it is reasonable.

User Wren
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