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Factorize the following expressions: (i) (x + 1)(x + 2)(x + 3)(x + 4) - 15, (ii) 4x(2x + 3)(2x - 1)(x + 1) - 54, (iii) (x - 3)(x + 2)(x + 3)(x + 8) + 56.

1. (i) (x² + 5x + 1)(x² + 5x + 9); (ii) (4x² + 4x - 9)(4x² + 4x + 6); (iii) (x² + 5x - 18)(x² + 5x + 4)
2. (i) (x² + 5x - 1)(x² + 5x - 11); (ii) (4x² + 4x + 9)(4x² + 4x - 6); (iii) (x² + 5x + 18)(x² + 5x - 4)
3. (i) (x² + 5x + 5)² - 15; (ii) (2x + 1)²(2x - 1)²; (iii) (x + 5)²(x - 1)²
4. (i) (x + 1)(x + 2)(x + 7); (ii) (2x - 3)(2x + 9); (iii) (x + 3)(x + 8)

Correct answer: (i) (x² + 5x + 1)(x² + 5x + 9); (ii) (4x² + 4x - 9)(4x² + 4x + 6); (iii) (x² + 5x - 18)(x² + 5x + 4)

Solution

(i) (x+1)(x+4) = x² + 5x + 4 and (x+2)(x+3) = x² + 5x + 6. Let t = x² + 5x. Product = (t+4)(t+6) - 15 = t² + 10t + 24 - 15 = t² + 10t + 9 = (t + 1)(t + 9) = (x² + 5x + 1)(x² + 5x + 9). (ii) 4x(2x+3)(2x-1)(x+1): group 2x(2x-1) = 4x² - 2x and (2x+3)*2(x+1) = (2x+3)(2x+2) = 4x² + 10x + 6. Actually arrange to share u = 4x² + 4x: (2x+3)(2x-1) = 4x² + 4x - 3 and 4x(x+1) = 4x² + 4x. Let u = 4x² + 4x. Product = u(u - 3) - 54 = u² - 3u - 54 = (u - 9)(u + 6) = (4x² + 4x - 9)(4x² + 4x + 6). (iii) (x-3)(x+8) = x² + 5x - 24 and (x+2)(x+3) = x² + 5x + 6. Let t = x² + 5x. Product = (t - 24)(t + 6) + 56 = t² - 18t - 144 + 56 = t² - 18t - 88 = (t - 22)(t + 4)... check: (t-22)(t+4) = t² - 18t - 88, correct. So = (x² + 5x - 22)(x² + 5x + 4). The provided option lists (x² + 5x - 18) which differs; the correct factorization is (x² + 5x - 22)(x² + 5x + 4), and (i),(ii) match the chosen option.

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