Multiply: (a) (m + 2)(m – 2), (b) (3y + 4)(3y – 4), (c) (b -6 +), and (d) (tª – 6u)(tª + 6u). To find the product of each pair of binomials, we will use the special-product rule for the sum and difference of the same two terms. This approach is faster than using the FOIL method. (a) (m + 2) and (m - 2) are the sum and difference of the same two terms, m and 2. (m + 2)(m – 2) = m2 22 The square of the first term, m. The square of the second term, 2. = m2 -O (b) (3y + 4) and (3y - 4) are the sum and difference of the same two terms, 3y and 4. (3y + 4)(3y – 4) = (3y)? 42 The square of the first term, 3y. The square of the second term, 4. |y² - O %3D (c) By the commutative property of multiplication, the special-product rule can be written with the factor containing the – symbol first. That is (A – B)(A + B) = A² - B?. Since (o-금) and (o + 금) are the difference and sum of the same two terms, b and 2/3, we have (금)" +o %3D The square of the first term, b. The square of the second term, 2/3. = b2 -O (d) (tª - 6u) and (t + 6u) are the difference and sum of the same two terms, tª and 6u. (e4 – 6u)(t + 6u) = (4)2 (6u)² The square of the first term, tª. The square of the second term, 6u. = t8 - Multiply: (a) (b + 7)(b-7), (b) (2m + 8)(2m- 8), (c) (s-흥)(s + 흥), and (d) (c³ + 3d)(c³ - 3d). (a) (b) (c) (d)