Section 4.5 Factoring Special Products All Binomials 1) Sum and Difference of cubes 2) Difference of Squares You must be able to recognize this list of cubes. 1, 8, 27, 64, πππ 125. How can you come up with this list on your own? (1)(1)(1) = π, (2)(2)(2) = π, (3)(3)(3) = ππ, 1. Difference of cubes / Sum of cubes: π¨π β π©π OR (4)(4)(4) = ππ, (5)(5)(5) = πππ π¨π + π©π A Difference/Sum of cubes is a binomial where both terms are perfect cubes. A difference of cubes always factors using the formula (π¨ β π©)(π¨π + π¨π© + π©π ) where A is the cube root of the first term and B is the cube root of the second term. A sum of cubes always factors using the formula (π¨ + π©)(π¨π β π¨π© + π©π ) where A is the cube root of the first term and B is the cube root of the second term. Example 1: Factor the following polynomials. π) 8π₯ + 27π¦ b) 5π₯ β 5 c) 128π₯ π¦ β 250π¦ You must be able to recognize this list of squares. 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 How can you come up with this list on your own? (1)(1) = π, (2)(2) = 4, (3)(3) = π, (6)(6) = ππ, (7)(7) = ππ, (8)(8) = ππ, (4)(4) = 1π, (5)(5) = ππ (9)(9) = ππ, (10)(10) = πππ 2. Difference of squares: π¨π β π©π A Difference of Squares is a binomial where both terms are perfect squares and there is a β-β sign in between. A difference of squares always factors as conjugates. (π¨ β π©)(π¨ + π©) where A is the square root of the first term and B is the square root of the second term. *NOTE: SUM of squares is PRIME. π¨π + π©π cannot be factored down into smaller parts. Example 2: Factor the following polynomials. a) 6π₯ β 24 d) 81π₯ β 1 b) 3π₯ π¦ β 27π₯ π¦ c) 16π₯ + 25 e) 81π₯ β 16π¦ f) 4π₯ + 12π₯ β 9π₯ β 27
© Copyright 2025 Paperzz