Find the product. Simplify your answer. (2r - 3)(2r + 3) ______

Special Case Identification: Which special case applies here? (2r - 3)(2r + 3) is in the form of (a - b)(a + b). Special case: (a - b)(a + b) = a^2 - b^2Values of a and b: Identify the values of a and b. Compare (2r - 3)(2r + 3) with (a - b)(a + b). a = 2r b = 3Difference of Squares Application: Apply the difference of squares to expand (2r - 3)(2r + 3). (a - b)(a + b) = a^2 - b^2 (2r - 3)(2r + 3) = (2r)^2 - (3)^2Simplification: Simplify (2r)^2 - (3)^2. (2r)^2 - (3)^2 = (2r * 2r) - (3 * 3) = 4r^2 - 9

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Find the product. Simplify your answer. $\newline$ $(2r - 3)(2r + 3)$

View step-by-step help

Home

Math Problems

Algebra 1

Multiply two binomials: special cases

Full solution

Q. Find the product. Simplify your answer.

\newline

(2r - 3)(2r + 3)

Special Case Identification: Which special case applies here? $\newline$ $(2r - 3)(2r + 3)$ is in the form of $(a - b)(a + b)$ . $\newline$ Special case: $(a - b)(a + b) = a^2 - b^2$
Values of $a$ and $b$ : Identify the values of $a$ and $b$ . Compare $(2r - 3)(2r + 3)$ with $(a - b)(a + b)$ . $a = 2r$ $b = 3$
Difference of Squares Application: Apply the difference of squares to expand $(2r - 3)(2r + 3)$ . $\newline$ $(a - b)(a + b) = a^2 - b^2$ $\newline$ $(2r - 3)(2r + 3) = (2r)^2 - (3)^2$
Simplification: Simplify $(2r)^2 - (3)^2.$ $(2r)^2 - (3)^2 = (2r \times 2r) - (3 \times 3)$ $= 4r^2 - 9$