Verified answer

x = 2 is a zero means, (x-2) is a factor:

by inspection:

x^4-x^3+3x^2-5x-10 = (x-2)(x^3 + x^2 + 5x + 5)

then by inspection, x = -1 is a root of the second factor, so

(x^3 + x^2 + 5x + 5) = (x+1)(x^2 + 5) = (x+1)[x-isqrt(5)][x+isqrt(5)]

x^4-x^3+3x^2-5x-10 = (x-2)(x+1)[x-isqrt(5)][x+isqrt(5)]

roots are:

x = 2, x = -1, x = +- sqrt(-5) = +- isqrt(5)

Using synthetic division and divide out the 2. This does give a 0 remainder and the result is x^3 + x^2 + 5x + 5

factoring by grouping:

x^2(x + 1) + 5(x + 1)

(x^2 + 5)(x + 1)

setting = 0 and solving for x, you get

x = +/- sqrt5 and x = -1 as the other three zeros