Question

(x+1)(x+2)(x-5)(x-6)-144=0

Answer 1

$(x + 1) \times (x + 2) \times (x - 5) \times (x - 6) - 144 = 0 \\ ( {x}^{2} + 2x + x + 2) \times (x - 5) \times (x - 6) - 144 = 0 \\ ( {x}^{2} + 3x + 2) \times (x - 5) \times (x - 6) - 144 = 0 \\ ( {x}^{3} - 5 {x}^{2} + 3 {x}^{2} - 15x + 2x - 10) \times (x - 6) - 144 = 0 \\ ( {x}^{3} - 2 {x}^{2} - 13x - 10) \times (x - 6) - 144 = 0 \\ {x}^{4} - {6}^{3} -2 {x}^{3} + 12 {x}^{2} - 13 {x}^{2} + 78 {x}^{2} - 10x + 60 - 144 = 0 \\ {x}^{4} - 8 {x}^{3} - {x}^{2} + 68x - 84 = 0 \\ {x}^{4} - 2 {x}^{3} - 6 {x}^{3} + 12 {x}^{2} - 13 {x}^{2} + 26x + 42x - 84 = 0 \\ {x}^{3} \times (x - 2) - 6 {x}^{2} \times (x - 2) - 13x \times (x - 2) + 42(x - 2) = 0 \\(x - 2) \times ( {x}^{3} - 6 {x}^{2} - 13x + 42) = 0 \\(x - 2) \times ( {x}^{3} - 2 {x}^{2} - 4 {x}^{2} + 8x - 21x + 42) = 0 \\ (x - 2) + ( {x}^{2} \times (x - 2) - 4x \times (x - 2) - 21(x - 2)) = 0 \\ (x - 2) \times (x - 2) \times ( {x}^{2} - 4x - 21) = 0 \\(x - 2) \times (x - 2) \times ( {x}^{2} + 3x - 7x - 21) = 0 \\ (x - 2) \times (x - 2) \times (x \times (x + 3) - (x + 3)) = 0 \\(x - 2) \times (x - 2) \times (x + 3) \times (x - 7) = 0 \\ {(x - 2)}^{2} \times (x + 3) \times (x - 7) = 0 \\ {(x - 2)}^{2} = 0 \\ x + 3 = 0 \\ x - 7 = 0 \\ x = 2 \\ x = - 3 \\ x = 7$

Answer 2

Answer:

Step-by-step explanation:

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