I can't always quickly simplify a linear factor from the expression as is.
For example:
2x³ - 3x² - 5x + 6
... But, I notice that the free-term has factors 1,2,3,6... Using these factors as roots one could substitute them as x into the remainder expression from the synthetic or long division algorithm such as:
2x³ - 3x² - 5x = -6
2(1)³ - 3(1)² - 5(1) = -6
I have a hunch that there must be a better and quicker way than this so please enlighten me.
Is there anything I'm obviously missing here?
更新1:
What I'm asking sorry, is there a way that is not guessing, when the expression will not factor on it's own?
更新2:
Also, I'm curious what it is about the free coefficient's factors that make this work... I'm also noticing at least some time these factors only work as the negative values, which are not directly the numbers you get from the free term's factors... Also I'm curious as to whether it has anything to do with the free term being negative?
