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Multiplying Pairs of Brackets and Simplifying

Learn how to expand and simplify expressions involving multiplying pairs of brackets using the Box Method, FOIL Method, and the Separation Method. Examples and exercises provided.

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Multiplying Pairs of Brackets and Simplifying

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  1. Multiplying Pairs of Brackets and Simplifying EXPANDING

  2. This means to multiply out a pair, or more, of brackets e.g.(ax + b) (cx + d).There are 2 ways of expanding brackets. They are: • The Box Method • The FOIL Method

  3. The Box Method Multiply the individuals components together (2a + 3)(4a + 2)

  4. The Box Method (2a + 3)(4a + 2)

  5. We now have four separate values so (2a+3) (4a+2) = 8a2 + 12a + 4a +6 This can be simplified to give us our final answer (2a+3) (4a+2) = 8a2 + 16a+6

  6. Exercise • (x + 2)(x + 1) • (x + 8)(x + 2) • (x + 3)(x + 3) • (x + 4)(x – 5) • (x – 9)(x + 2) • (x + 8)(x – 3) • (x – 1)(x – 6) • (x – 2)(x + 8) • (x + 2)(5 – x) • (8 – x)(1 – x) • x2 + 3x + +2 • x2 +10x + 16 • x2 + 6x + 9 • x2 – x - 20 • x2 – 7x - 18 • x2 + 5x - 24 • x2 – 7x + 6 • x2 + 6x – 16 • -x2 + 3x + 10 • x2 – 9x + 8

  7. The FOIL method FOIL stands for: FIRST, OUTER, INNER, LAST And refers to the order in which the values are multiplied. For Example : (x + 3) (x - 5) FIRST, xxx = x2 OUTER, x x-5 = -5x INNER, +3xx = 3x LAST +3 x-5= -15 SIMPLIFY (x + 3) (x - 5) = x2 -2x -15

  8. Example 2 : (2x - 3) (4x - 5) FIRST, 2xx4x = 8x2 OUTER, 2x x-5 = -10x INNER, -3x4x = -12x LAST -3 x-5 = +15 SIMPLIFY (2x - 3) (4x - 5) = 8x2 -22x +15

  9. Exercise • (x + 5) (x + 5) • (2x + 3) (5x – 4) • (x – 3) (x – 3) • (3 + x) (3 + x) • (2x – 9) (2x + 9) • (x + 9) (x – 2) • (5x – 1) (6x + 2) • (3x – 7) (2x – 3) • (2 + x) (x – 5) • (8 + 2x) (1 + x) • x2 + 10x +25 • 10x2 + 7x – 12 • x2 – 6x + 9 • x2 + 6x + 9 • 4x2 – 81 • x2 + 7x – 18 • 30x2 + 4x – 2 • 6x2 – 23x + 21 • x2 – 3x – 10 • 2x2 + 10x + 8

  10. The Separation Method Separating the brackets will often make life much easier. For example (x + 5) (2x – 3) = x(2x – 3) +5(2x-3) Now we can multiply out the separate brackets to obtain x(2x – 3) = 2x2 -3xand 5(2x-3) = 10x -15 Adding these will give 2x2 -3x + 10x -15 = 2x2 + 7x -15

  11. Example 2 (x2 + 2x - 5) (2x – 3) = x2(2x – 3) + 2x(2x-3) – 5(2x-3) Now we can multiply out the separate brackets to obtain x2(2x – 3) = 2x3 - 3x2 2x(2x – 3) = 4x2 – 6x and -5(2x-3) = -10x +15 Adding these will give 2x3 -3x2 +4x2 -6x - 10x +15 = 2x3 + x2 - 16x +15

  12. Exercise • (x + 4) (x + 4) • (x + 3) (5x – 4) • (x + 3) (x – 3) • (3 + x) (3 – x) • (2x + 9) (2x + 9) • (2x + 9) (4x – 2) • (6x – 2) (x + 1) • (2x – 7) (3x – 3) • (2 + 3x) (x2 – 4x) • (8 + x + x2) (1 + 5x) • x2 + 8x +16 • 5x2 + 11x – 12 • x2 – 9 • 9 – x2 • 4x2 + 36x + 81 • 8x2 + 32x – 18 • 6x2 + 4x – 2 • 6x2 – 27x + 21 • 3x3 – 10x2 – 8x • 5x3 + 6x2 + 41x + 8

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