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Math Help - Help with Polynomial, Factorizing and Equations (semi-urgent)

  1. #1
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    Help with Polynomial, Factorizing and Equations (semi-urgent)

    I need help with the following problems.

    1.) Factor: y + 1000y^4


    2.) Use the general factoring strategy to help factor this polynomial:

    x^3 + 2x^2 + 2x^3 - 6x -15 + 3x^2 + 5

    3.) Use the general factoring strategy (and the zero product property) to help solve this equation:

    x^2 - 20x = -x^2 - 9x - 5

    4.) Some of these trinomials are factorable, some are prime. Factor those that can be factored:

    A.] 6 + 5t - 6t^2

    B.] 6 + 5t + 6t^2

    C.] 6 - 5t + 6t^2

    D.] 6 - 5t - 6t^2

    5.)

    Use factoring to identify the zero's of this polynomial: 3x^5 - 30x^3 + 27x


    Any help in explaining how you do these problems is greatly appreciated! I missed the first couple days of class due to illness and came back swamped, and lost as far as what to do.

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by zagsfan20 View Post
    I need help with the following problems.

    1.) Factor: y + 1000y^4
    y(1 + 1000y^3)

    You'll need to recognize it's of the form x^3 + y^3; see formula

    Thus,

    y*(10y + 1)*(100y^2 - 10y + 1)
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  3. #3
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    Quote Originally Posted by zagsfan20 View Post

    2.) Use the general factoring strategy to help factor this polynomial:

    x^3 + 2x^2 + 2x^3 - 6x -15 + 3x^2 + 5
    Combine like-terms:

    3x^3 + 5x^2 - 6x - 10

    Factor:

    (x^2 - 2)*(3x + 5)
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  4. #4
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    Quote Originally Posted by zagsfan20 View Post
    3.) Use the general factoring strategy (and the zero product property) to help solve this equation:

    x^2 - 20x = -x^2 - 9x - 5
    x^2 - 20x = -x^2 - 9x - 5; bring all terms to one side

    2x^2 - 11x + 5 = 0

    Several ways to solve this. Let's change it up and use the quadratic formula; I do not know the "general factoring strategy" that your teacher showed you.

    x = [11 +/- sqrt(121 - 4*2*5)]/(2*2)

    x = [11 +/- sqrt(81)]/4

    x = [11 +/- 9]/4

    x = 20/4 AND x = 2/4

    Thus, x = 5 and x = 1/4.

    We could have also done this by factoring, which I personally find to be quicker.
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    Quote Originally Posted by zagsfan20 View Post
    4.) Some of these trinomials are factorable, some are prime. Factor those that can be factored:

    A.] 6 + 5t - 6t^2

    B.] 6 + 5t + 6t^2

    C.] 6 - 5t + 6t^2

    D.] 6 - 5t - 6t^2
    A.] Can be factored;

    (-2t + 3)*(3t + 2)

    B.] CANNOT be factored.

    C.] CANNOT be factored.

    D.] Can be factored;

    Very similar to part A, although there is a sign change.

    (-2t + 3)(3t - 2)
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  6. #6
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    Quote Originally Posted by zagsfan20 View Post
    I need help with the following problems.

    1.) Factor: y + 1000y^4


    2.) Use the general factoring strategy to help factor this polynomial:

    x^3 + 2x^2 + 2x^3 - 6x -15 + 3x^2 + 5

    3.) Use the general factoring strategy (and the zero product property) to help solve this equation:

    x^2 - 20x = -x^2 - 9x - 5

    4.) Some of these trinomials are factorable, some are prime. Factor those that can be factored:

    A.] 6 + 5t - 6t^2

    B.] 6 + 5t + 6t^2

    C.] 6 - 5t + 6t^2

    D.] 6 - 5t - 6t^2

    5.)

    Use factoring to identify the zero's of this polynomial: 3x^5 - 30x^3 + 27x


    Any help in explaining how you do these problems is greatly appreciated! I missed the first couple days of class due to illness and came back swamped, and lost as far as what to do.

    Thanks in advance!
    You posted this in another thread!! Please refer to rule #1, here. I'll leave it to you to decide if it was Aftershock or me that wasted our time.

    -Dan
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  7. #7
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    Quote Originally Posted by zagsfan20 View Post
    5.)[/B]

    Use factoring to identify the zero's of this polynomial: 3x^5 - 30x^3 + 27x
    3x^5 - 30x^3 + 27x = 0

    3x*(x^4 - 10x^2 + 9) = 0

    3x*[(x^2)^2 - 10(x^2) + 9)]

    3x*[(x^2 - 4x + 3)*(x^2 + 4x + 3)]

    Separate above two terms and factor each;

    (x - 3)*(x - 1) and (x + 1)*(x + 3)

    Thus:

    3x*(x - 3)*(x - 1)*(x + 1)*(x + 3) = 0

    Thus, the zeros are:

    x = -3, x = -1, x = 0, x = 1, x = 3
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  8. #8
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    Quote Originally Posted by topsquark View Post
    You posted this in another thread!! Please refer to rule #1, here. I'll leave it to you to decide if it was Aftershock or me that wasted our time.

    -Dan
    Meh. What a shame. Well, Dan, our results agree .
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    Thank you to both of you guys!

    Much appreciated.
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by AfterShock View Post
    Meh. What a shame. Well, Dan, our results agree .
    Always a nice thing to have happen.

    -Dan
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