Let f(x) = x^3-2x^2-5x+6f(x)=x3−2x2−5x+6
First note that the sum of the coefficients is 00, so x=1x=1 is a zero of f(x)f(x)...
f(1) = 1-2-5+6 = 0f(1)=1−2−5+6=0
Divide f(x)f(x) by (x-1)(x−1) to find:
f(x) = x^3-2x^2-5x+6 = (x-1)(x^2-x-6)f(x)=x3−2x2−5x+6=(x−1)(x2−x−6)
Then x^2-x-6 = (x-3)(x+2)x2−x−6=(x−3)(x+2)
So the other two roots of f(x) = 0f(x)=0 are x=3x=3 and x=-2x=−2
Alternatively, use the rational roots theorem to know that any rational roots of f(x) = 0f(x)=0 must be of the form p/qpq where pp, qq are integers, q != 0q≠0, pp a factor of the constant term 66 and qq a factor of the coefficient 11 of the term (x^3x3) of highest degree. So the only possible rational roots are:
+-1±1, +-2±2, +-3±3 and +-6±6
Try each to find f(1) = 0f(1)=0, f(-2) = 0f(−2)=0 and f(3) = 0f(3)=0.
