I'll see if I can get you to an understanding of the remainder theorem...
Let's take the function f(x) = x^2 + 5x - 6
substitute 1 for x. What do you get? 0
Divide the function by (x-1). What's the remainder? 0
f(x) = x^2 +5x -6 can be factored as (x-1)(x+6)
That's why a value of 1 for x causes it to be zero.
Now, let's review something else for a moment. Look at the graph of f(x) = x^2. Now, look at the graph of f(x) = x^2 + 1. Now, look at the graph of f(x) = x^2 +7. In the first graph, the vertex was at the origin. In the second graph, every point was "lifted up" one unit. In the 3rd graph, the original graph was shifted 7 units vertically. So, adding a number or subtracting a number at the end of a function serves to shift it up or down that many units.
Now, let's look at f(x) = x^2 + 5x - 6 again. We noticed that when x = 1, the value of this function is 0. That's because (x-1) is a factor of the function. Thus, if you divided x^2 + 5x - 6 by (x-1), you'd get a remainder of 1.
Now, for fun, let's add 17 to this function. f(x) = x^2 + 5x - 6 + 17, or
f(x) = x^2 + 5x + 11. Plug in 1 for x, the same x value as last time. Hmmm... f(x) = 17. That shouldn't be a surprise; by adding 17 to the function, we took an original point on the function (1,0), and raised it 17 units to a new point: (1,17)
Now, let's see what happens when we divide f(x)=x^2 + 5x +11 by (x-1). We get a remainder of 17! This also isn't a coincidence. Look at the quotient: (x + 6). By dividing, we have figured out that
x^2 + 5x + 11 can be written as (x-1)(x+6) + a remainder of 17.
Thus, f(x) = (x+1)(x+6) + 17
This shows us that this new function (with +11 for the constant) is simply the old function raised 17 units.
Now, start with a function of the order you desire; one that has a root of zero somewhere. Now, shift (translate) your function by adding a value to the constant. Check your answer.
