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Math Help - Factoring Questions - Help!

  1. #1
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    Factoring Questions - Help!

    We were recently given a worksheet with some problems on it and I'm struggling with a few of the problems. Can anyone lead me in the direction or show me how to do some of these factoring problems?

    Factor Fully.

    1) 28s^2 - 8st - 20t^2

    2) y^2 - (r - n)^2

    3) 10x^2 + 38x + 20

    4) y^5 - y^4 + y^3 - y^2 + y - 1

    I'd greatly appreciate any help given for any of these problems. Thank you!
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  2. #2
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    Quote Originally Posted by Jeavus View Post
    1) 28s^2 - 8st - 20t^2

    2) y^2 - (r - n)^2

    3) 10x^2 + 38x + 20

    4) y^5 - y^4 + y^3 - y^2 + y - 1
    1) 28s^2-8st-20t^2=4(7s^2-2st-5t^2), it remains to factorice 7s^2-2st-5t^2=7s^2-(7st-5st)-5t^2, can you take it from there?

    2) Use a^2-b^2=(a+b)(a-b)

    3) Similar as 1)

    4) Factorice by grouping
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  3. #3
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    Quote Originally Posted by Jeavus View Post
    We were recently given a worksheet with some problems on it and I'm struggling with a few of the problems. Can anyone lead me in the direction or show me how to do some of these factoring problems?

    Factor Fully.

    1) 28s^2 - 8st - 20t^2
    28s^2 - 8st - 20t^2

    First rule, factor any constant out of the whole thing that you can:
    = 4(7s^2 - 2st - 5t^2)

    Now, since 7 and 5 are prime all we need to do is arrange the 7 and 5 in some way to multiply out correctly. Trial and error gives me:
    = 4(7s + 5t)(s - t)

    Quote Originally Posted by Jeavus View Post
    2) y^2 - (r - n)^2
    Remember a^2 - b^2 = (a + b)(a - b)?
    Let a = y, b = r - n.
    y^2 - (r - n)^2 = (y + (r - n))(y - (r - n)) = (y + r - n)(y - r + n)

    Quote Originally Posted by Jeavus View Post
    3) 10x^2 + 38x + 20
    10x^2 + 38x + 20

    = 2(5x^2 + 19x + 10)
    This can't be factored any further than this.

    Quote Originally Posted by Jeavus View Post
    4) y^5 - y^4 + y^3 - y^2 + y - 1
    This is a fun one, isn't it? Factor by grouping:
    y^5 - y^4 + y^3 - y^2 + y - 1

    = (y^5 - y^4 + y^3) + (- y^2 + y - 1)

    = y^3(y^2 - y + 1) - (y^2 - y + 1)

    Notice that both terms in parenthesis are the same, so we can factor again:
    = (y^3 - 1)(y^2 - y + 1)

    Now, the second factor can't be factored any further, but the first term can:
    a^3 - b^3 = (a - b)(a^2 + ab + b^2)

    So
    y^5 - y^4 + y^3 - y^2 + y - 1 = (y^3 - 1)(y^2 - y + 1) = (y - 1)(y^2 + y + 1)(y^2 - y + 1)

    -Dan
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  4. #4
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    1) 28s^{2} - 8st - 20t^{2}
    The trick here is to factor out 4 first.

    Then we have 4(7s^{2}-2st-5t^{2})

    Now, look at the middle term in the parentheses.

    Rewrite -2st as 5st-7st:

    4(7s^{2}+5st-7st-5t^{2})

    4((7s^{2}+5st)-(7st+5t^{2}))

    4(s(7s+5t)-t(7s+5t))

    4(s-t)(7s+5t)
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  5. #5
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    Thanks so much for all the help! That all made perfect sense.

    I have a few more questions that I am struggling with if you'd be so kind as to direct me in the correct path:

    1) 3v^2 - 11v - 10

    Personally I don't think this one can be factored, as nothing multiplies to (-30) and adds to (-11).

    2) h^3 + h^2 + h + 1

    3) 9(x + 2y + z)^2 - 16(x - 2y + z)^2

    All help will be greatly appreciated.
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  6. #6
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    They all have the same idea, try it by your own!
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    I'm completely stumped with the first one. =/

    I've tried various grouping methods with the second.

    And I'm assuming there is some kind of trinomial factoring but again, I'm lost as to finding the light.

    Any hints? :|
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  8. #8
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    Here's where I'm stuck for the 2nd one:

    h^3 + h^2 + h + 1
    = h(h^2 + h + 1) + 1

    Now I need two numbers that multiply to 1, but also add to 1 - which doesn't make sense to me. =/

    EDIT AGAIN:

    Solved this one!

    Anyone have any ideas about the other two?
    Last edited by Jeavus; September 7th 2007 at 09:34 PM.
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  9. #9
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    Quote Originally Posted by Jeavus View Post
    Thanks so much for all the help! That all made perfect sense.

    I have a few more questions that I am struggling with if you'd be so kind as to direct me in the correct path:

    1) 3v^2 - 11v - 10
    Are you sure that you have the signs right? Could it be:

    v^2-11v+10?

    RonL
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  10. #10
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    Yes, I have the signs right.

    Un-factorable, right?
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  11. #11
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    It has real nasty roots, but you could factorice it finding its roots.
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  12. #12
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    Krizalid/anyone else, do you have any hints towards the third one?

    9(x + 2y + z)^2 - 16(x - 2y + z)^2
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  13. #13
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    Quote Originally Posted by Jeavus View Post
    9(x + 2y + z)^2 - 16(x - 2y + z)^2
    9\left( {x + 2y + z} \right)^2 - 16\left( {x - 2y + z} \right)^2 = \left[ {3\left( {x + 2y + z} \right)} \right]^2 - \left[ {4\left( {x - 2y + z} \right)} \right]^2

    Does that make sense?

    Now apply the identity a^2-b^2=(a+b)(a-b)
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  14. #14
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    9(x + 2y + z)^2 - 16(x - 2y + z)^2
    = [3(x + 2y + z)]^2 - [4(x - 2y + z)]^2
    a^2 - b^2
    a = [3(x + 2y + z)]
    b = [4(x - 2y + z)]
    (a + b)(a - b)
    = ([3(x + 2y + z)] + [4(x - 2y + z)])([3(x + 2y + z)] - [4(x - 2y + z)])

    Where do I go from this stage?
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  15. #15
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    Quote Originally Posted by Jeavus View Post
    Yes, I have the signs right.

    Un-factorable, right?
    No real factors

    RonL
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