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

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Presentation on theme: "Revision - Algebra I Binomial Product Using Distributive Law to Expand / Remove Brackets. By I Porter."— Presentation transcript:

1 Revision - Algebra I Binomial 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 x 33(7 - 4)= 3 x x 4 5 x 5 = x 3= =259= 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 Example: Expand (a + 4) ( a + 2) Geometrical Method Algebraic Method a+ 4 a + 2 aa + 4()() Area = The area of the large rectangle is equal to the sum of the areas of all the smaller rectangles. a2a2 + 4a + 2a+ 8 (a+4)(a+2) = a 2 + 4a + 2a + 8 = a 2 + 6a + 8 aa+ 2)(+ 4a+ 2() = a 2 + 2a + 4a + 8 = a 2 + 6a + 8 By the distributive law: F+O+I+L. method of distributive law. FOIL is a mnemonic - a memory jogger - indicating order of multiplication (don’t forget negative signs). O = a x +2 = +2 a I = +4 x a = +4 a L = +4 x +2 = +8 So, (a + 4)(a + 2) = a 2 + 4a + 2a + 8 = a 2 + 6a + 8 The algebraic method(s) are the preferred. (a + 4)(a + 2) = F = a x a = a 2 F = first terms O = outside terms I = inside terms L = last terms

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

5 Exercise: Remove the grouping symbols. a) 4y(3 - 2y) = b) -8x 2 (3 - 2x) = c) (x + 9)(3x + 4) = d) (x - 7)(2x - 5) = e) (3x + 2)(2x + 5) = f) (x + 12)(x - 12) = g) (3x + 2y) 2 = h) (7 - 2xy) 2 = i) (8x - 3y) (8x + 3y) = 12y - 8y -24x x 3 3x x x x x x + 10 x x xy + 4y xy + 4x 2 y 2 64x 2 - 9y 2 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) = = 3x 2 + x + 5 = 3x x + 21 = 2x x + 34 = 5x x + 6 = 26x


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