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Math Help - Factoring With Respect to the Integers

  1. #1
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    Factoring With Respect to the Integers

    I'm a little bit stuck on a few of my homework problems, and can't really find out how to do them in my text book. I don't really need help on both of them unless there're different ways to find the answer, I just need to learn how to solve them. Any help would be fantastic.
    Thanks in advance,
    Mackenzie

    "Completely factor the polynomial with respect to the integers"
    1. x2-9x-22
    2. 2x2-5x+2
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  2. #2
    Super Member bigwave's Avatar
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    Quote Originally Posted by mackenzie25 View Post
    I'm a little bit stuck on a few of my homework problems, and can't really find out how to do them in my text book. I don't really need help on both of them unless there're different ways to find the answer, I just need to learn how to solve them. Any help would be fantastic.
    Thanks in advance,
    Mackenzie

    "Completely factor the polynomial with respect to the integers"
    1. x2-9x-22
    2. 2x2-5x+2
    given your expressions do you mean

    1.  x^2-9x-22
    2. 2x^2-5x+2

    if so

    1. (x-11)(x+2)
    2. (2x-1)(x-1) or 2(x-2)(x-.5)

    there is no roots since these are not equations just expressions
    Last edited by bigwave; December 28th 2009 at 01:24 PM. Reason: more info
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  3. #3
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    That's definitely what I was going for, but how did you factor it?
    (Thanks! )
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  4. #4
    Super Member bigwave's Avatar
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    ax^2 + bx + c where a, b, c , are real numbers

     <br />
(mx+p)(nx+q)<br />
    where

     <br />
mn = a, pq = c<br />
    and

     <br />
pn + mq = b<br />
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  5. #5
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    (5)(7)=(3+2)(3+4)=3(3+4)+2(3+4)=3^2+3(4)+2(3)+2(4)

    =3^2+[3(4)+2(3)]+2(4)=35

    Such\ an\ alternative\ allows\ us\ to\ expand\ (x+a)(x+b)
    to\ x(x+b)+a(x+b)=x^2+(a+b)x+ab

    Also\ (5)(7)=(9-4)(9-2)=9^2+9(-4-2)+(-4)(-2)=35
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