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Algebra help

  • 29-05-2012 12:45pm
    #1
    GarIT
    Registered Users, Registered Users 2 Posts: 8,671 ✭✭✭


    I have a sample question here and I haven't seen the second part before and don't have a clue how to do it. Can anyone help? The question is

    Alpha and beta are roots of x^2-6x-2=0
    (a) find alpha x beta and alpha+beta
    (b) Factorise alpha^3 + beta^3

    How do I factorise that? Is it (a+b)((a+b)^2 -3ab)


Comments

  • #2
    finality
    Registered Users, Registered Users 2 Posts: 1,763 ✭✭✭finality


    (a + b)^3 - 3ab(a + b)

    Edit: though that's the same as what you said, so yup that's right! :P


  • #3
    leaveiton
    Closed Accounts Posts: 1,778 ✭✭✭leaveiton


    Is it not (a + b)(a^2 - ab + b^2) in the case of a^3 + b^3? Or is that the same? :P


  • #4
    finality
    Registered Users, Registered Users 2 Posts: 1,763 ✭✭✭finality


    leaveiton wrote: »
    Is it not (a + b)(a^2 - ab + b^2) in the case of a^3 + b^3? Or is that the same? :P

    In a situation like this you need to have only (a + b) or ab, as these are what you get from the equation (you know, like ab=c/a)

    using a and b here as alpha and beta.

    but both are correct. I mean if you did it your way you'd just need to take (a^2 + b^2) separately as (a + b)^2 - 2ab
    and it would work out fine, it's just an extra step.


  • #5
    ChemHickey
    Closed Accounts Posts: 3,479 ✭✭✭ChemHickey


    finality wrote: »
    leaveiton wrote: »
    Is it not (a + b)(a^2 - ab + b^2) in the case of a^3 + b^3? Or is that the same? :P

    In a situation like this you need to have only (a + b) or ab, as these are what you get from the equation (you know, like ab=c/a)

    using a and b here as alpha and beta.

    but both are correct. I mean if you did it your way you'd just need to take (a^2 + b^2) separately as (a + b)^2 - 2ab
    and it would work out fine, it's just an extra step.


    Some (my teacher included) think its an extra hassle to learn the other formula. I prefer it the separate formulas but it's up to you anyway.


  • #6
    leaveiton
    Closed Accounts Posts: 1,778 ✭✭✭leaveiton


    Ah okay, I get what you're saying. I think I remember seeing it that way in a book but we were never taught it. Thanks! :)


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  • #7
    finality
    Registered Users, Registered Users 2 Posts: 1,763 ✭✭✭finality


    ChemHickey wrote: »
    Some (my teacher included) think its an extra hassle to learn the other formula. I prefer it the separate formulas but it's up to you anyway.

    but you're still going to have to learn the a^2 + b^2 one. the a^3 + b^3 one is almost the same anyway :)


  • #8
    GarIT
    Registered Users, Registered Users 2 Posts: 8,671 ✭✭✭GarIT


    You don't even need to learn it if you look at it logically, it just works that (a+b)^3 -3ab(a+b) = a^3 + b^3. The way I remember it is that (a+b)^3 is the same as a^3 + b^3, but it just has an extra little bit, so you just make them equal to each other and take away the extra ab bit.

    I was thinking that factors are things you multiply together to make another number or equation. When I thought about it I thought that (a+b)^3 -3ab(a+b) is subtracting one number from the other and not a factor but that (a+b)((a+b)^2 -3ab) is a factor because you are multiplying two numbers.


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