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Revision - Algebra I Binomial Product

Revision - Algebra I Binomial Product. Using Distributive Law to Expand / Remove Brackets. By I Porter. Definitions. Grouping symbols (…..) can be expanded by using the Disributlive Law. This states that for any real numbers a, b and c : a(b ± c) = ab ± ac Examples:

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Revision - Algebra I Binomial Product

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  1. Revision - Algebra IBinomial Product Using Distributive Law to Expand / Remove Brackets. By I Porter

  2. Definitions Grouping symbols (…..) can be expanded by using the Disributlive Law. This states that for any real numbers a, b and c: a(b ± c) = ab ± ac Examples: 5(2+3) = 5 x 2 + 5 x 3 3(7 - 4) = 3 x 7 - 3 x 4 5 x 5 = 10 + 15 3 x 3 = 21 - 12 25 = 25 9 = 9 Binomial Products A binomial expression contains two (2) terms such as : (2 + n), (2x - 5), (3n - 5p) A binomial product is the multiplication of two such binomial expression: (2 + n)(2x - 5), (a - b) (a + b), (2n - 5)(3n + 2), (3p + 4)2. The product of two binomials can be obtained using two approaches: a) the Geometrical Approach - using area diagrams. b) the Algebraic Approach - using the distributive law.

  3. (a + 4)(a + 2), can be represented by the area of a rectangle with sides (a + 4) and (a + 2). Area of the rectangle = length x breadth ( a + 2 ) ( a + 2 ) ( a + 4 ) ( a + 2 ) Area = a + 4 F = first terms a a2 + 4a O = outside terms I = inside terms + 2 + 2a + 8 L = last terms Example: Expand (a + 4) ( a + 2) Algebraic Method By the distributive law: Geometrical Method (a + 4)(a + 2) = a + 4 = a2 + 2a + 4a + 8 = a2 + 6a + 8 The area of the large rectangle is equal to the sum of the areas of all the smaller rectangles. F+O+I+L. method of distributive law. FOIL is a mnemonic - a memory jogger - indicating order of multiplication (don’t forget negative signs). F = a x a = a2 O = a x +2 = +2 a I = +4 x a = +4 a L = +4 x +2 = +8 So, (a + 4)(a + 2) = a2 + 4a + 2a + 8 = a2 + 6a + 8 (a+4)(a+2) = a2 + 4a + 2a + 8 = a2 + 6a + 8 The algebraic method(s) are the preferred.

  4. x(5x - 2) - 3(5x - 2) -3 x 4x - - 3 x 5 x(x + 4) - 7 (x + 4) = 3x(3x + 4) + 4(3x + 4) x(2x - 1) + 5(2x - 1) Distributive Law Method 4) (5x - 2)(x - 3) = Examples: Remove the grouping symbols: = 5x2 -2x -15x + 6 1) -3(4x - 5) = = 5x2 -17x + 6 = -12x + 15 (3x + 4)(3x + 4) 5) (3x + 4)2 = 2) (x + 4) (x -7) = = x2 + 4x - 7x - 28 = 9x2 + 12x +12x + 16 = x2 -3x - 28 = 9x2 + 24x + 16 Special cases : Difference of two squares. 3) (2x - 1) (x + 5) = 6) (8 - x) (8 + x) = 8(8 - x) + x(8 - x) = 2x2 - x + 10x - 5 = 64 - 8x + 8x - x2 = 2x2 + 9x - 5 = 64 - x2

  5. j) (2x - 1) (x + 4) + (x - 3)2 k) (2x + 3)2 - (x + 4) (x - 3) l) (x - 3)2 + (x - 5)2 = m) (3x - 1)2 - (4x - 1)(x + 5) = n) (5x - 3) (5x + 3) - (4 - x) (4 + x) = Exercise: Remove the grouping symbols. a) 4y(3 - 2y) = 12y - 8y = 3x2 + x + 5 b) -8x2 (3 - 2x) = -24x2 + 16x3 c) (x + 9)(3x + 4) = 3x2 + 31x + 36 = 3x2 + 11x + 21 d) (x - 7)(2x - 5) = 2x2 - 19x + 35 e) (3x + 2)(2x + 5) = 6x2 + 19x + 10 = 2x2 - 16x + 34 f) (x + 12)(x - 12) = x2 - 144 g) (3x + 2y)2 = 9x2 + 12xy + 4y2 = 5x2 - 25x + 6 h) (7 - 2xy)2 = 49 - 28xy + 4x2y2 = 26x2 - 25 i) (8x - 3y) (8x + 3y)= 64x2 - 9y2

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