Question

(x+2)(x-5)(x-6)(x+1)=144

Answer 1

$(x + 2)(x - 6)(x - 5)(x + 1) = 144 \\ ( {x}^{2} - 4x - 12)( {x}^{2} - 4x - 5) = 144 \\ let \: \: {x}^{2} - 4x = y \\ then \\ (y - 12)(y - 5) = 144 \\ {y}^{2} - 17y - 84 = 0 \\ (y - 21)(y + 4) = 0 \\ y = 21 \: \: or \: \: - 4 \\ \\ i. \: \: if \: y = 21 \\ {x}^{2} - 4x - 21 = 0 \\ (x - 7)(x + 3) = 0 \\ x = 7 \: \: or \: \: - 3 \\ ii. \: \: if \: y = - 4 \\ {x}^{2} - 4x + 4 = 0 \\ {(x - 2)}^{2} = 0 \\ x = 2 \\ \\ therefore \: \: solution \: \: is \\x = 2 \: \: or \: \: - 3 \: \: or \: \: 7$

Answer 2

Answer:

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