Find the product. Simplify your answer. (3q+1)(3q-1)

Identify special case: Identify the special case for the product (3q+1)(3q-1). This product is in the form of (a+b)(a-b), which is a difference of squares. Special case: (a+b)(a-b) = a^2 - b^2Identify values of a and b: Identify the values of a and b. Compare (3q+1)(3q-1) with (a+b)(a-b). a = 3q b = 1Apply difference of squares formula: Apply the difference of squares formula to expand (3q+1)(3q-1). (a+b)(a-b) = a^2 - b^2 (3q+1)(3q-1) = (3q)^2 - (1)^2Simplify expression: Simplify (3q)^2 - (1)^2. (3q)^2 - (1)^2 = (3q * 3q) - (1 * 1) = 9q^2 - 1

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Find the product. Simplify your answer.

(3q+1)(3q-1)

Find the product. Simplify your answer. $\newline$ $(3 q+1)(3 q-1)$

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Algebra 1

Multiply two binomials: special cases

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Q. Find the product. Simplify your answer.

\newline

(3 q+1)(3 q-1)

Identify special case: Identify the special case for the product $(3q+1)(3q-1)$ . This product is in the form of $(a+b)(a-b)$ , which is a difference of squares. Special case: $(a+b)(a-b) = a^2 - b^2$
Identify values of $a$ and $b$ : Identify the values of $a$ and $b$ . Compare $(3q+1)(3q-1)$ with $(a+b)(a-b)$ . $a = 3q$ $b = 1$
Apply difference of squares formula: Apply the difference of squares formula to expand $(3q+1)(3q-1)$ . $\newline$ $(a+b)(a-b) = a^2 - b^2$ $\newline$ $(3q+1)(3q-1) = (3q)^2 - (1)^2$
Simplify expression: Simplify $(3q)^2 - (1)^2.$ $(3q)^2 - (1)^2 = (3q \times 3q) - (1 \times 1)$ $= 9q^2 - 1$