Find the product. Simplify your answer. (2k + 1)(2k - 1) ______

Identify a and b: We are given the expression (2k + 1)(2k - 1) and need to find its product. This expression is in the form of (a + b)(a - b), which is a difference of squares. The special case for the difference of squares is: (a + b)(a - b) = a^2 - b^2.Apply difference of squares: Identify the values of a and b. Compare (2k + 1)(2k - 1) with (a + b)(a - b). a = 2k b = 1Expand and simplify: Apply the difference of squares formula to expand (2k + 1)(2k - 1). (a + b)(a - b) = a^2 - b^2 (2k + 1)(2k - 1) = (2k)^2 - (1)^2Simplify (2k)^2 - (1)^2. (2k)^2 - (1)^2 = (2k * 2k) - (1 * 1) = 4k^2 - 1

Home

Math Problems

Algebra 1

Multiply two binomials: special cases

Full solution

Q. Find the product. Simplify your answer.

\newline

(2k + 1)(2k - 1)

Identify $a$ and $b$ : We are given the expression $(2k + 1)(2k - 1)$ and need to find its product. $\newline$ This expression is in the form of $(a + b)(a - b)$ , which is a difference of squares. $\newline$ The special case for the difference of squares is: $(a + b)(a - b) = a^2 - b^2$ .
Apply difference of squares: Identify the values of $a$ and $b$ . $\newline$ Compare $(2k + 1)(2k - 1)$ with $(a + b)(a - b)$ . $\newline$ a = $2k$ $\newline$ b = $1$
Expand and simplify: Apply the difference of squares formula to expand $(2k + 1)(2k - 1)$ . $\newline$ $(a + b)(a - b) = a^2 - b^2$ $\newline$ $(2k + 1)(2k - 1) = (2k)^2 - (1)^2$
Expand and simplify: Apply the difference of squares formula to expand $(2k + 1)(2k - 1)$ . $\newline$ $(a + b)(a - b) = a^2 - b^2$ $\newline$ $(2k + 1)(2k - 1) = (2k)^2 - (1)^2$ Simplify $(2k)^2 - (1)^2$ . $\newline$ $(2k)^2 - (1)^2 = (2k \cdot 2k) - (1 \cdot 1)$ $\newline$ $= 4k^2 - 1$