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Math Help - Remainder and Factor Theorem

  1. #1
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    Remainder and Factor Theorem

    I know how to use both the remainder theorem and the factor theorem, but I'm not sure how to use them in relation to this problem:
    When a polynomial is divided by x-p, the remainder is p^3.
    When a polynomial is divided by x-q, the remainder is q^3.

    Find the remainder when the polynomial is divided by (x-p)(x-q).
    Any help would be greatly appreciated, even if it's just a few hints to point me in the right direction.
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  2. #2
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    Quote Originally Posted by CUBE View Post
    I know how to use both the remainder theorem and the factor theorem, but I'm not sure how to use them in relation to this problem:
    When a polynomial is divided by x-p, the remainder is p^3.
    When a polynomial is divided by x-q, the remainder is q^3.

    Find the remainder when the polynomial is divided by (x-p)(x-q).
    Any help would be greatly appreciated, even if it's just a few hints to point me in the right direction.
    Hello, maybe you are looking for this?: http://www.mathhelpforum.com/math-he...lynomials.html
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    I don't understand the method used in that thread. Could somebody else explain it better?
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  4. #4
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    Quote Originally Posted by CUBE View Post
    I know how to use both the remainder theorem and the factor theorem, but I'm not sure how to use them in relation to this problem:
    When a polynomial is divided by x-p, the remainder is p^3.
    When a polynomial is divided by x-q, the remainder is q^3.

    Find the remainder when the polynomial is divided by (x-p)(x-q).
    Any help would be greatly appreciated, even if it's just a few hints to point me in the right direction.
    I can't think of a less complex method to use. Perhaps you just need a little more explanation red_dog's method.

    You are given a polynomial f(x). You know that when you divide f(x) by x - p the remainder is p^3. So you know that
    f(x) = (x - p)g_1(x) + p^3
    by the division theorem, where g_1(x) is some polynomial. By the remainder theorem (or just plain substitution) we also see that
    f(p) = (p - p)g_1(p) + p^3 = p^3

    Now, f(x) divided by x - q leaves a remainder of q^3. So by exactly the same line of reasoning we have that
    f(x) = (x - q)g_2(x) + q^3 \implies f(q) = q^3
    where g_2(x) is another polynomial.

    Now we wish to divide f(x) by (x - p)(x - q). Now, we are dividing by a quadratic polynomial. When we do this in general we are not left with a constant remainder, we are left with a remainder polynomial of one degree less than the divisor: we are left with a first degree polynomial as the remainder. So f(x) must be of the form:
    f(x) = (x - p)(x - q)g_3(x) + (ax + b)
    where g_3(x) is another polynomial.

    Possibly you would be more comfortable with this format?
    \frac{f(x)}{(x - p)(x - q)} = g_3(x) + \frac{ax + b}{(x - p)(x - q)}
    This is the same statement.

    Plugging x = p and x = q into this equation gives us:
    f(p) = (p - p)(p - q)g_3(p) + (ap + b) = ap + b
    and
    f(q) = (q - p)(q - q)g_3(q) + (aq + b) = aq + b

    But we already know that f(p) = p^3 and f(q) = q^3 so we have the system of equations:
    ap + b = p^3
    aq + b = q^3

    Now solve for a and b:
    a = p^2 + pq + q^2
    b = -pq(p +q)

    (Note that if p = q we have no solution for a and b, so the problem really needed so specify that p \neq q.)

    So when f(x) is divided by (x - p)(x - q) we get a remainder of (p^2 + pq + q^2)x - pq(p + q).

    If you want a concrete example of all this, consider f(x) = x^3 + 2x^2 - 2x - 4 and p = -1 and q = 2.

    -Dan
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