Applying the rational roots theorem, we find that any rational roots of the given expression will be of the form p/qpq where pp is a divisor of -5−5 and qq is a positive divisor of 11. Then, the possible rational roots are +-1±1 and +-5±5.
Trying these out, we find that (-1)^3+3(-1)^2-3(-1)-5 = 0(−1)3+3(−1)2−3(−1)−5=0 Thus x-(-1) = x+1x−(−1)=x+1 is a factor of x^3+3x^2-3x-5x3+3x2−3x−5
Dividing, we get
x^3+3^2-3x-5 = (x + 1)(x^2 + 2x -5)x3+32−3x−5=(x+1)(x2+2x−5)
To find the remaining roots, we can simply apply the quadratic formula
ax^2+bx+c = 0 => x = (-b+-sqrt(b^2-4ac))/(2a)ax2+bx+c=0⇒x=−b±√b2−4ac2a
to obtain the roots of x^2+2x-5x2+2x−5 as
(-2+-sqrt(2^2-4(1)(-5)))/(2(1)) = -1 +- sqrt(6)−2±√22−4(1)(−5)2(1)=−1±√6
The the final set of roots (zeros) of the expression is
{-1, -1+sqrt(6), -1-sqrt(6)}{−1,−1+√6,−1−√6}
