2014/028) solve for real a (a+5)(a+4)(a+3)^2(a+2)(a+1) = 360

putting a + 3 =t we gen

(t+2)(t+1)t^2(t-1)(t-2) = 360

or rearranging the terns and multiplying we get

t^2(t^2-1)(t^2-4) = 360

putting t^2= x we get

x(x-1)(x – 4) = 360 (1)(

so x^3-5x^2 + 4x – 360 = 0

as 360 = 9 * 8 * 5 so from (1) x = 9 is a root

so we get factoing ( x – 9)(x^2 + 4x + 40) = 0





so x = 9 => a + 3 = 3 or – 3 so a = 0 or – 6

x^2 + 4x + 40 = 0 gives complex solutions