Question

solve(x+1)(x+2)(x+3)(x+4)=120​

Answer 1

Four solutions, two real and two complex.

Note that (x+1)(x+4) = x^2+5x+4

(x+2)(x+3) = x^2+5x+6

If we put y = x^2+5x+5, then x^2+5x+4 =y-1 and x^2+5x+6 = y+1

Then the equation becomes (y-1)(y+1)=120 or y^2–1=120 or y^2=121

First solution y1 = 11

x^2+5x+5 = 11 gives x^2+5x-6 = 0 x1=1, x2=-6

The second value of y gives

x^2+5x+5 = -11 or x^2+5x+16=0. This produces two complex solutions

x3=-5/2+isqrt(39)/2, x4=-5/2-isqrt(39)/2

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