By the rational root theorem any rational roots of f(x) = 0 must be of the form p/q where p and q are integers, q != 0, p a divisor of the constant term -10 and q a divisor of the coefficient 1 of the term of highest degree x^4.
So the only possible rational roots are:
+-1, +-2, +-5, +-10
We find:
f(1) = 1 + 1 + 1 - 9 - 10 = -16
f(-1) = 1 - 1 + 1 + 9 - 10 = 0
f(2) = 16 + 8 + 4 - 18 - 10 = 0
So at least x = -1 and x = 2 are roots of f(x) = 0 and (x+1) and (x-2) are factors of f(x).
Divide f(x) by (x+1)(x-2) = x^2-x-2 to find:
f(x) = x^4+x^3+x^2-9x-10
= (x^2-x-2)(x^2+2x+5)
Using the quadratic formula, the roots of x^2+2x+5 = 0 are:
x = (-2+-sqrt(2^2-(4xx1xx5)))/(2*1)
=(-2+-sqrt(4-20))/2
=(-2+-sqrt(-16))/2
=-1+-2i
