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Math Help - Need help with a couple of questions

  1. #1
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    Need help with a couple of questions

    Any help would be greatly appreciated.

    1. When an expression f(x) is divided by 3x+1, the quotient is 2x-3 and the remainder is x+5. What is f(x)?

    2. Factor the following:

    (a) x^12n - y^20n
    (b) x^2 + 4xy +4y^2 - m^2 - 6mn - 9n^2 [EDIT - SOLVED?]: (x+2y)^2 - (m+3n)^2
    (c) 1/3x^-3 y^6 -1/4x^4y^3
    Last edited by tmac20; September 25th 2011 at 09:21 AM.
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  2. #2
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    Re: Need help with a couple of questions

    1. Suppose when x is divided by 3, the quotient is 5 and the remainder is 1. Can you find x? Also, it is essential to know the definition of the remainder. Can you write it?

    2.

    (a) Use the formula a^2 - b^2 = (a - b)(a + b).
    (b) Use the formula a^2 + 2ab+b^2=(a+b)^2 and the one in (a).
    (c) You can either factor out x^{-3}y^6 or use the formula in (a).
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  3. #3
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    Re: Need help with a couple of questions

    Quote Originally Posted by emakarov View Post
    1. Suppose when x is divided by 3, the quotient is 5 and the remainder is 1. Can you find x? Also, it is essential to know the definition of the remainder. Can you write it?

    2.

    (a) Use the formula a^2 - b^2 = (a - b)(a + b).
    (b) Use the formula a^2 + 2ab+b^2=(a+b)^2 and the one in (a).
    (c) You can either factor out x^{-3}y^6 or use the formula in (a).
    Thanks, I understand #2 now, but I'm still having trouble with #1. (I haven't done dividing by hand since I was in Grade 6.) How do I find f(x)?
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  4. #4
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    Re: Need help with a couple of questions

    The quotient and remainder when a natural number x is divided by y are natural numbers q and r such that x = qy + r and 0\le r<y. If you know y, q and r, you can find x from the equation above.
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  5. #5
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    Re: Need help with a couple of questions

    Quote Originally Posted by emakarov View Post
    The quotient and remainder when a natural number x is divided by y are natural numbers q and r such that x = qy + r and 0\le r<y. If you know y, q and r, you can find x from the equation above.
    But how do I find y?
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  6. #6
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    Re: Need help with a couple of questions

    The number y in my post is what you divide by. In the original problem about polynomials, this is given.
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  7. #7
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    Re: Need help with a couple of questions

    Quote Originally Posted by emakarov View Post
    The number y in my post is what you divide by. In the original problem about polynomials, this is given.
    Thanks, one more question - how do you factor out x^-3 y^6?
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  8. #8
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    Re: Need help with a couple of questions

    one more question - how do you factor out x^-3 y^6?
    As usual. When you factor out a from m\cdot a - n\cdot b, you get a(m - n\cdot b/a). Here a = x^{-3} y^6 and b = x^4y^3, so you need to find (x^4y^3)/(x^{-3} y^6). Remember that x^k/x^l = x^{k-l} and take care when l is negative.

    There are several ways to factor m\cdot a - n\cdot b. This also equals b(m\cdot a/b-n) and (\sqrt{m}\sqrt{a}+\sqrt{n}\sqrt{b})(\sqrt{m}\sqrt{  a}-\sqrt{n}\sqrt{b}).
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