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**Expanding two brackets**

Look at this algebraic expression: (3 + t)(4 – 2t) This means (3 + t) × (4 – 2t), but × is not used in algebra. To expand or multiply out this expression, multiply every term in the second bracket by every term in the first bracket. Teacher notes In this example, multiply everything in the second bracket by 3 and then everything in the second bracket by t. This can be written as 3(4 – 2t) + t(4 – 2t). As pupils become more confident, they can leave this intermediate step out. (3 + t)(4 – 2t) = 3(4 – 2t) + t(4 – 2t) This is a quadratic expression. = 12 – 6t + 4t – 2t2 = 12 – 2t – 2t2 2

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**Using the grid method Teacher notes**

Demonstrate as many examples as is necessary for pupils to understand the method. Once pupils are happy with the method they may be able to multiply out brackets without using a grid. 3

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**Expanding two brackets**

The product of two linear expressions can be expanded in fewer steps. For example, (x – 5)(x + 2) = x2 + 2x – 5x – 10 = x2 – 3x – 10 Notice that –3 is the sum of –5 and 2… …and that –10 is the product of –5 and 2. Teacher notes Point out that for any expression in the form: (x + a)(x + b), where a and b are fixed numbers, the expanded expression will have an x with a coefficient of a + b and the number at the end will be a × b. 4

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**Matching quadratic expressions 1**

Teacher notes Select a bracketed expression and ask a volunteer to find its corresponding expansion. 5

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**Matching quadratic expressions 2**

Teacher notes Select a bracketed expression and ask a volunteer to find its corresponding expansion. 6

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**Squaring expressions Expand and simplify: (2 – 3a)2**

(2 – 3a)2 = (2 – 3a)(2 – 3a) = 2(2 – 3a) – 3a(2 – 3a) = 4 – 6a – 6a + 9a2 = 4 – 12a + 9a2 7

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Squaring expressions It can be seen that there is a pattern relating a squared expression and its form once expanded and simplified. (a + b)2 = a2 + 2ab + b2 The first term squared… …plus 2 × the product of the two terms… …plus the second term squared. For example, (3m + 2n)2 = 9m2 + 12mn + 4n2 8

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**Squaring expressions Teacher notes**

Any of the terms in the expansion can be hidden or revealed to practice squaring expressions. 9

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**The difference between two squares**

Expand and simplify: (2a + 7)(2a – 7) (2a + 7)(2a – 7) = 2a(2a – 7) + 7(2a – 7) = 4a2 – 14a + 14a – 49 = 4a2 – 49 When simplifying, the two middle terms cancel out. This is the difference between two squares. (a + b)(a – b) = a2 – b2 10

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**The difference between two squares**

Teacher notes This animation demonstrates geometrically why a2 – b2 = (a – b)(a + b). The first two steps in the animation demonstrate a square of area b2 being subtracted from a square of area a2 to give a shape of area a2 – b2. The shape of area a2 – b2 is then rearranged in to a rectangle of width a + b and height a – b to show that its area is equal to (a – b)(a + b). a2 – b2 is therefore equal to (a – b)(a + b). 11

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**The difference between two squares**

Teacher notes Select a bracketed expression and ask a volunteer to find its corresponding expansion. 12

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Pendants of gold Here are two pendant patterns which are to be made out of gold. The white squares are holes. 2x + 2 x 1 x x + 1 1 3x + 2 2x 5 2x Teacher notes The amount of gold that pendant one uses can be calculated as: (2x + 2)(3x + 2) – 4x2 – 10x. This simplifies to: 2x2 + 4. The amount of gold that pendant two uses can be calculated as: 2x(x + 1) – 4. This simplifies to: 2x2 +2x – 4. 2x2 +2x – 4 – (2x2 + 4) = 2x – 8. When x = 4, 2x – 8 = 0 (same amount of gold) They do not always use the same amount of gold. When x = 3, 2x – 8 = –2. In this instance, pendant 2 uses less gold. When x = 5, 2x – 8 = 2. In this instance, pendant 2 uses more gold. This could be extended with pupils working in pairs comparing two different pendants of their own design. For what value of x do they both use the same amount of gold? Do they always use the same amount of gold? 13

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