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The remainder theorem

  1. Feb 1, 2007 #1
    Find a polynomial of degree three that when divided by x - 2 has a remainder of 3. You will really have to think on this one. Hint: Work backwards!

    ok heres the thing i've tried i've looked at other problems but I can barely work problems forward, backwards...well your talking to me here my math skills are beyond horrible! my very sad attempt at this goes as follows


    could someone just help at like walking me through it or tell me the first step i'll take anything.
  2. jcsd
  3. Feb 1, 2007 #2
    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.
  4. Feb 2, 2007 #3


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    Science Advisor

    (x-2) times any second degree polynomial you want, plus 3!

    That's all you need to do.
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