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Expand an equation - sum and product

  1. May 24, 2014 #1
    1. The problem statement, all variables and given/known data
    I have been sitting here for the last hour trying to figure it out but I can't seem to be able to find what I'm doing wrong.
    I need to expand an equation.


    2. Relevant equations
    a2 - a - 3


    3. The attempt at a solution
    a2 - 1a - 3
    The product is -3 and the sum -1.

    -3 * 1 = -3
    -1 * 3 = -3.
    Both products are -3.
    I can't seem to be able to find anything though of which the sum is also -1.
    My sums are always either -2 or 2.
     
  2. jcsd
  3. May 24, 2014 #2

    Dick

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    That's not an equation and you aren't trying to expand it. What you appear to be trying to do is factor a quadratic. That quadratic doesn't factor in integers. Are you trying to solve a^2-a-3=0?
     
  4. May 24, 2014 #3

    SteamKing

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    If you are trying to write the expression a^2 - a - 3 as the product of two monomial factors where everything has nice integer coefficients and such, well, good luck. There aren't any such factors. Sorry you killed an hour on this effort.

    In the future, if the factors of the constant term can't produce the coefficient of the linear term, stop, don't waste your time like you did here.
     
  5. May 24, 2014 #4
    English isn't my native language so I mixed up the 2 things. I am indeed trying to factor a quadratic.
    The full question is to simplify it as far as possible and make use of factoring a quadratic.
    The full expression is: [itex]\frac{a^2+2a-8}{a^2-a-2}[/itex] / [itex]\frac{a^3+4a^2}{a^2-a-3}[/itex]

    I managed to figure the first one out.
    [itex]\frac{a^2+2a-8}{a^2-a-2}[/itex]

    a2+2a-8
    (a + 4)(a - 2)
    Which is indeed a2+ 2a - 8

    a2 - 1a - 2
    (a - 2)(a + 1)
    a2 + 1a - 2a - 2
    a2 - 1a - 2

    [itex]\frac{a^2+2a-8}{a^2-a-2}[/itex] = [itex]\frac{(a+4)(a-2)}{(a-2)(a+1)}[/itex]
    Now I can remove the (a-2) because they both have it.
    So [itex]\frac{a^2+2a-8}{a^2-a-2}[/itex] = [itex]\frac{(a+4)}{(a+1)}[/itex]

    I can't figure out the second part of the expression.

    As my first problem is I have no clue how to do the a3+4a2
    I decided to instead already solve the bottom. Which I also can't seem to figure out.
     
  6. May 24, 2014 #5

    Dick

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    a3+4a2 isn't quadratic but it has a simple factorization. Factor the common factor of ##a^2## out. ##a^2-a-3## won't factor. You'll just have to leave that as it is.
     
  7. May 24, 2014 #6

    HallsofIvy

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    The equation [itex]a^2- a- 3= 0[/itex] has roots [tex]a= \frac{1+\sqrt{13}}{2}[/tex] and [itex]a= \frac{1- \sqrt{13}}{2}[/itex]. That means that [itex]a^2- a- 3[/itex] factors as
    [tex]\left(a- \frac{1}{2}- \frac{\sqrt{13}}{2}\right)\left(a- \frac{1}{2}+ \frac{\sqrt{13}}{2}\right)[/tex]

    That's probably doesn't help you!
     
  8. May 25, 2014 #7
    I think that
    [itex]a^2- a- 3[/itex]
    is much simpler than the roots so I'll just leave that as it is but I appreciate your effort to help me.

    I came up with:
    [itex]\frac{a+4}{a+1}[/itex] / [itex]\frac{a^2(a+4)}{a^2-a-3}[/itex]
     
  9. May 25, 2014 #8
    Thank you! I took a look into that and it seems to work out.
     
  10. May 25, 2014 #9

    SammyS

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    Of course, that's far from being simplified. I hope you realize that.
     
  11. May 25, 2014 #10
    I do. :P
    I just like to respond to other people so they know I have read their comments and let them know that I appreciate their efforts.
     
  12. May 25, 2014 #11

    SammyS

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    There's no problem with that, in fact you are to commended for giving the feedback. Some folks just disappear after being helped, leaving no indication that they finally 'get it'.

    However, often the person starting the thread gives the final result after being helped.
     
  13. May 26, 2014 #12
    Woops, I thought I did. I forgot apparently.
    My final answer on how to simplify it by factoring is
    [itex]\frac{a+4}{a+1}[/itex] / [itex]\frac{a^2(a+4)}{a^2-a-3}[/itex]
    The right part of the expression is as a result of common factoring. Both share a2
    Instead of factoring the last one out with the roots, I decided to leave it as it was because of having to simplify.
    The (a+4)(a+1) explanation is because a2+2a-8 equals (a+4)(a-2) and a2-2-a equals (a-2)(a+1). I can eliminate the (a-2) because they both have it. What remains is (a+4) and (a+1). Therefore a+4 / a+1
     
  14. May 26, 2014 #13

    SammyS

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    [itex]\displaystyle {\frac{a+4}{a+1}\ }/{\ \frac{a^2(a+4)}{a^2-a-3}}[/itex] can be simplified to a single rational function. I'm quite certain that's what you are supposed to do.

    There's still more '"cancelling" to be done after that.
     
  15. May 26, 2014 #14
    I have tried to do that but I kept failing on how to do so.
    I have tried asking other people, they couldn't figure it out either.
    As I already had to send my answer in, I just went with what I had.
    Mind telling me how to do simplify it further for future purposes?
     
  16. May 26, 2014 #15

    SammyS

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    I would write it as a compound (a.k.a. complex fraction) first:

    [itex]\displaystyle \frac{\displaystyle \frac{a+4}{a+1}}{\displaystyle \ \frac{a^2(a+4)}{a^2-a-3}\ }[/itex]

    Then multiply the numerator & denominator of the main fraction by something to cancel both sub-denominators or else by something to cancel the entire denominator of the main fration.
     
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