Algebra
posted by Yvonne .
(x^3+2x^24x8)/(x^416)x(3x^2+8x+5)/3x^2+11x+10)

factor each part
here is the hardest part
x^3 + 2x^2  4x  8 , using grouping
= x^2(x+2)  4(x+2)
= (x^2  4)(x+2)
= (x+2)(x2)(x+2) or (x2)(x+2)^2
so...
(x^3+2x^24x8)/(x^416)x(3x^2+8x+5)/3x^2+11x+10)
= (x2)(x+2)^2 /((x2)(x+2)(x^2 + 4)) * (x+1)(3x+5)/((3x+5)(x+2))
= (x+1)/(x^2 + 4) , x ≠ ±2, 5/3
check by picking any value of x other than the restricted values
e.g. let x = 1
in original:
= (9)/(15)*(16/24) = 2/5
value of (x+1)/(x^2+4) = 2/5