Question 173481
From a table of what?


If you have two x values that each correspond to a y value of 0, it is a simple process.


Let's say that you have two values for x, {{{x=a}}} and {{{x=b}}}, such that your table of function values has


{{{x=a}}}, {{{f(a)=0}}} and 


{{{x=b}}}, {{{f(b)=0}}}


Then you can say {{{(x-a)(x-b)=0}}} and then multiply the binomials to get {{{x^2-(a+b)x+ab=0}}}