How do you find the zeros of #y=x^3-2x^2-11x+12#?
The zeros are
Sum of coefficients shortcut
Note that the sum of the coefficients is
#1-2-11+12 = 0#
#x^3-2x^2-11x+12 = (x-1)(x^2-x-12)#
Fishing for factors
To factor the remaining quadratic we want a pair of factors of
#x^2-x-12 = (x-4)(x+3)#
So the zeros of the given cubic are
So the given example cubic proved to be not too difficult, but what might we do if the sum of coefficients shortcut didn't apply?
Rational zeros theorem
Given any polynomial with integer coefficients in standard form, any rational zeros are expressible in the form
In our example, any rational zeros would take the form
#+-1, +-2, +-3, +-4, +-6, +-12#
We could try substituting each in turn until we find a zero, then divide the polynomial by the corresponding factor as before.