Question 124775
Given the cubic equation: {{{ax^3+bx^2+cx+d=0}}}, the only possible rational roots are of the form x = factor of d/factor of a or x= -(factor of d/factor of a), so by trial and error find s such that {{{(x-s)}}} divides the original equation by polynomial long division without a remainder.


For your problem, a = 1 and d = 10.  The possible factors of d are 1, 2, and 5.  1 is the only factor of a.  So if a rational root exists, it must be ±1, ±2, or ±5.


When I was working this out, I started with s = 1, dividing  {{{x^3 -4x^2 -7x +10}}} by {{{x-1}}}.  I got lucky, or so I thought.


The quotient after performing the polynomial long division, was {{{x^2-3x-10}}}


Now all that remains is to factor {{{x^2-3x-10}}}.  {{{-5*2=-10}}} and {{{-5+2=-3}}}, so our factors are {{{(x-5)}}} and {{{(x+2)}}}.  Turns out that luck wasn't a factor -- I had a 50-50 chance of finding the correct factor at the start, knowing that I had ±1, ±2, or ±5 to choose from.


Therefore: {{{x^3 -4x^2 -7x +10=green((x-1)(x+2)(x-5))}}}


I'll let you multiply it out to check the answer.