MHB Solve Equation III: $(a+5)(a+4)(a+3)^2(a+2)(a+1)=360$

[anemone]

Solve in complex solutions the equation $(a+5)(a+4)(a+3)^2(a+2)(a+1)=360$.

 

[kaliprasad]

Solve in complex solutions the equation $(a+5)(a+4)(a+3)^2(a+2)(a+1)=360$.

I find 2 real solutions

Spoiler

putting a + 3 =t we get

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

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

$x^2 + 4x + 40 = 0 => (x + 2)^2 = - 36$
or x = - 2 +/- 6i

this shall give 4 roots for a by taking the square root