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Math Help - just checking that I got these right

  1. #1
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    Thumbs up just checking that I got these right

    Just want to double check that I am doing these right. Not looking for an answer.


    5p^4q - 20q^3 should = 5pq(p^3-4q^2) right?


    72x^2 - 2y^2 should = (2y-3x) (2y-36x) right?

    Thanks so much!
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  2. #2
    Senior Member chella182's Avatar
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    Quote Originally Posted by tpoma View Post
    Just want to double check that I am doing these right. Not looking for an answer.


    5p^4q - 20q^3 should = 5pq(p^3-4q^2) right?


    72x^2 - 2y^2 should = (2y-3x) (2y-36x) right?

    Thanks so much!
    Well the first one is wrong because if you multiply out your answer you get 5p^{4}q-20pq^{3}, which isn't what you have on the left-hand side. In other words, you can't take a p outside the bracket.

    I don't think the second one is right either because multiplying that out you'd get 4y^{2}-6xy-72xy+108x^{2} or 4y^{2}-78xy+108x^{2} which again isn't what you have on the left-hand side.
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  3. #3
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    ok re-do

    5p^4q - 20q^3 should have been 5q(p^4-4q^2)

    Will have to work on the second one.....
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  4. #4
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    Quote Originally Posted by tpoma View Post
    Just want to double check that I am doing these right. Not looking for an answer.


    5p^4q - 20q^3 should = 5pq(p^3-4q^2) right?
    Find your greatest common factor over your two terms

    5p^4q-20q^3

    The common factor is 5q. Factor that out and you have:

    5q(p^4-4q^2)


    Quote Originally Posted by tpoma View Post
    72x^2 - 2y^2 should = (2y-3x) (2y-36x) right?

    Thanks so much!
    Find your GCF again. That would be 2. Factor that out.

    72x^2-2y^2=2(36x^2-y^2)

    Inspecting the binomial, you should notice that it is the difference of two squares and factors in this manner:

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

    So, finally we have:

    72x^2-2y^2=2(36x^2-y^2)={\color{red}2(6x-y)(6x+y)}

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  5. #5
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    Thumbs up

    Quote Originally Posted by masters View Post
    Find your greatest common factor over your two terms

    5p^4q-20q^3

    The common factor is 5q. Factor that out and you have:

    5q(p^4-4q^2)




    Find your GCF again. That would be 2. Factor that out.

    72x^2-2y^2=2(36x^2-y^2)

    Inspecting the binomial, you should notice that it is the difference of two squares and factors in this manner:

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

    So, finally we have:

    72x^2-2y^2=2(36x^2-y^2)={\color{red}2(6x-y)(6x+y)}


    I wish I had your brains!!! This is so fustrating to me!! Thank you once again! I will get it one day hopefully lol.....
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  6. #6
    A riddle wrapped in an enigma
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    Quote Originally Posted by tpoma View Post
    I wish I had your brains!!!

    The Brain -- is wider than the Sky --
    For -- put them side by side --
    The one the other will contain
    With ease -- and You -- beside --

    The Brain is deeper than the sea --
    For -- hold them -- Blue to Blue --
    The one the other will absorb --
    As Sponges -- Buckets -- do --

    The Brain is just the weight of God --
    For -- Heft them -- Pound for Pound --
    And they will differ -- if they do --
    As Syllable from Sound --

    Emily Dickenson (w.1955)
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  7. #7
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    Ok got one more for you, and hopefully the last one....lol


    c^2 - 3cd - 18d^2

    is this correct?

    3(cd + 6d^2 - c^2) = 3c(d+2d^2 - c) gee I hope so!!
    Last edited by mr fantastic; December 14th 2008 at 03:03 AM. Reason: Remove a potentially offensive reference to a deity
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  8. #8
    Senior Member chella182's Avatar
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    You can't take a three out because there isn't a 3 in every term (the c^2 term). You can't take a c out either because there's no c in the d^{2} term.
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  9. #9
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    Quote Originally Posted by chella182 View Post
    You can't take a three out because there isn't a 3 in every term (the c^2 term). You can't take a c out either because there's no c in the d^{2} term.

    Ok so.....once again I am not able to do anything with this problem because it is not factorable....there is not a c, d, or a number in each term, so again my answer would be not able to factor. Why would my professor put these on here to be factored completely if you can not factor at all?
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  10. #10
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    Quote Originally Posted by tpoma View Post
    Ok got one more for you, and hopefully the last one....lol


    c^2 - 3cd - 18d^2

    is this correct?

    3(cd + 6d^2 - c^2) = 3c(d+2d^2 - c)
    I hate to burst your bubble, but 3 \ \ nor \ \ c is a common factor of your trinomial expression.

    You must factor this as a trinomial. There are different techniques people use to do this, but here I might suggest the "guess and check" method.

    If factorable, a trinomial will be the product of two binomials. So, let's start there.

    c^2 - 3cd - 18d^2=(\ \ \ \ \ \ \ \ )(\ \ \ \ \ \ \ \ )

    Now, factoring the first term is easy:

    c^2 - 3cd - 18d^2=(c \ \ \ \ \ \ )(c \ \ \ \ \ \ )

    Now, look at the last term coefficient -18. We need to find two factors that multiply together to make -18 and add together to make your middle term coefficient -3. This is the guess and check part.

    Try various factors of -18

    -3 \cdot +6 = -18
    -3 + 6 = 3

    This pair is no good because they don't add to -3

    -6\cdot +3 = -18
    -6 + 3 = -3

    This pair works because they add to -3

    So, let us finish filling in the two binomial factors. By the way, don't forget that there's a d^2 attached to that 18 we just factored.

    c^2 - 3cd - 18d^2=(c -6d)(c +3d)

    There you have it.
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  11. #11
    Senior Member chella182's Avatar
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    Actually...

    (c-6d)(c+3d)

    is the answer. If you multiply out those brackets you should get what you started with.

    Darn, got beaten to it I am so slow at factorising stuff like that.
    Last edited by mr fantastic; December 14th 2008 at 03:04 AM. Reason: Removed a (potentially) mild offensive word
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  12. #12
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    OH brother! You make that look so easy! Thanks!
    Last edited by mr fantastic; December 14th 2008 at 03:05 AM. Reason: Removed a potentially offensive word
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  13. #13
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    Last one I PROMISE!

    Last one I PROMISE!

    is this correct.....just a plain yes or no answer.

    49x^2 + 140x + 100

    answer should be......(7x + 10) (7x + 10) right???? Please say yes..lol
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  14. #14
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    Quote Originally Posted by tpoma View Post
    Last one I PROMISE!

    is this correct.....just a plain yes or no answer.

    49x^2 + 140x + 100

    answer should be......(7x + 10) (7x + 10) right???? Please say yes..lol
    Yes!
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  15. #15
    Member great_math's Avatar
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    Quote Originally Posted by tpoma View Post
    Ok got one more for you, and hopefully the last one....lol


    c^2 - 3cd - 18d^2

    is this correct?

    3(cd + 6d^2 - c^2) = 3c(d+2d^2 - c) god I hope so!!
    you can check my multiplying it again if you are not sure..... that's the way to check yourself
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