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Find the products for each special case:

6. (2x + 1)^2

(2x + 6)(2x – 6)
(x – 8)^2

1 Answer

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Final answer:

The products of the expressions (2x + 1)^2, (2x + 6)(2x – 6), and (x – 8)^2 are determined by applying the binomial square and difference of squares patterns, yielding 4x^2 + 4x + 1, 4x^2 - 36, and x^2 - 16x + 64 respectively.

Step-by-step explanation:

The student is asking for the products of special case binomial and trinomial expressions. These include the square of a binomial and the product of a sum and a difference, both of which are special patterns in algebra.

To find the product of (2x + 1)^2, you need to apply the binomial square pattern (a + b)^2 = a^2 + 2ab + b^2. Thus, (2x + 1)*(2x + 1) = 4x^2 + 4x + 1.

The product of (2x + 6)(2x – 6) is a difference of squares, which follows the pattern (a + b)(a - b) = a^2 - b^2. Therefore, (2x + 6)(2x – 6) = 4x^2 - 36.

Lastly, to find the product of (x – 8)^2, you again use the binomial square pattern. So, (x - 8)(x - 8) = x^2 - 16x + 64.

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