2x8 3x2 + 5x 2 – (x 2)(x 5) – – 2x x2 6x + 9 (x 3)2 is an identity because it is true for all values of x, but – – x2 5x + 10 = (x 3)2 is an equation because it is only true when x = -1. – – x2 + 5x + 4 = (x + 4)(x + 1) so the length is x + 4 and the width is x + 1, giving a perimeter of 4x + 10 = 2(2x + 5). Area = (2x 2)(2x 2) = 2x2 4x + 2 – – – Factorising by the number of squares gives 64(x2 4x + 4), then factorising again to – find the length of the side of each square gives x2 4x + 4 = (x 2)(x 2). Side length – – – is x 2 cm. – Extension a, b, a + b, a + 2b, 2a + 3b, 3a + 5b, 5a + 8b, 8a + 13b, 13a + 21b, 21a + 34b. Sum of the first ten terms is 55a + 88b = 11(5a + 8b).
2x8 3x2 + 5x – 2 (x – 2)(x – 5) 2x x2 – 6x + 9 (x – 3)2 is an identity because it is true for all values of x, but x2 – 5x + 10 = (x – 3)2 is an equation because it is only true when x = -1. x2 + 5x + 4 = (x + 4)(x + 1) so the length is x + 4 and the width is x + 1, giving a perimeter of 4x + 10 = 2(2x + 5). Area = (2x – 2)(2x – 2) = 2x2 – 4x + 2 Factorising by the number of squares gives 64(x2 – 4x + 4), then factorising again to find the length of the side of each square gives x2 – 4x + 4 = (x – 2)(x – 2). Side length is x – 2 cm. Extension a, b, a + b, a + 2b, 2a + 3b, 3a + 5b, 5a + 8b, 8a + 13b, 13a + 21b, 21a + 34b. Sum of the first ten terms is 55a + 88b = 11(5a + 8b).
266584-foundation-topic-check-in-6.01-algebraic-expressions.docx