Question 902623
if f(a) = r, then the point (a,r) must be on the graph of the equation.


when you divide an equation by x - a, then this is the same as finding f(a) with r being the remainder of the division.


if you divide the polynomial by x+3, this is the same as finding f(-3) and the result of that will be the remainder of 2.


this means that the point (-3,2) must be on the graph.


an example will show you what i mean.


assume the equation is x^2 + x - 4


if you divide that equation by x+3, you will get 2 as a remainder.


if you evaluate f(x) = x^2 + x - 4 at f(-3), you will get (-3)^2 - 3 - 4 = 9 - 7 = 2


the point (-3,2) must be on the graph of the equation.


you can see if that's true by graphing x^2 + x - 4 and x = -3 and y = 2 and see if they intersect on the graph of that equation.


the graph looks like this:


the intersection of the vertical line at x = -3 and the horizontal line at y = 2 with the graph of the equation x^2 + x - 4 is at the point (-3,2).


{{{graph(600,600,-10,10,-10,10,x^2 + x - 4,2,75(x+3))}}}


this is all part of the remainder theorem which you can read at the following link.


<a href = "http://www.purplemath.com/modules/remaindr.htm" target = "_blank">http://www.purplemath.com/modules/remaindr.htm</a>