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Math Help - Need Help! factoring

  1. #1
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    Need Help! factoring

    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
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by zagsfan20 View Post
    1.) Factor: y + 1000y^4
    y + 1000y^4

    The only common factor to both terms is a y, so:
    y + 1000y^4 = y(1 + 1000y^3)

    Now recall that a^3 + b^3 = (a + b)(a^2 + ab + b^2). Thus:
    y + 1000y^4 = y(1 + 1000y^3) = y(1 + 10y)(1 + 10y + 100y^2)

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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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
    I've been trying to figure out why this is split up the way it is, but I can't figure it. Anyway, here's my approach:

    x^3 + 2x^2 + 2x^3 - 6x -15 + 3x^2 + 5 = 3x^3 + 5x^2 - 6x -10

    I'm going to rearrange the order of the terms here:
    3x^3 + 5x^2 - 6x - 10 = (3x^3 - 6x) + (5x^2 -10)

     = 3x(x^2 - 2) + 5(x^2 - 2) = (3x + 5)(x^2 - 2)

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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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
    I'm not sure what is meant by the "general factoring strategy." You'll have to ask your instructor about that.

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

    2x^2 - 11x + 5 = 0

    Start listing the factors of 2 x 5 = 10 and see which add to make -11.
    Factors.....Sum
    1, 10 ........11
    2, 5............7
    etc.

    I get that -1 x -10 = 10 and -1 + -10 = -11.
    Thus:
    2x^2 - 11x + 5 = 0

    2x^2 - x - 10x + 5 = 0

    (2x^2 - x) - (10x - 5) = 0 <-- Watch the negative signs here.

    x(2x - 1) - 5(2x - 1) = 0

    (x - 5)(2x - 1) = 0

    Now, if ab = 0 then one or both of a and b must be 0. Thus:
    x - 5 = 0 or 2x - 1 = 0

    x = 5 or x = 1/2

    -Dan
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  5. #5
    Forum Admin topsquark's Avatar
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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
    For all of these, just to make sure we are working with the same format, write the polynomial in standard order: with the higher degree terms to the left.

    So A becomes: -6t^2 + 5t + 6

    Now use the same method I showed you in 3:
    I want two factors of -6 x 6 = -36 that add to 5.
    Factors......Sum
    1,-36.........-35
    2,-18.........-16
    etc.

    I get that 9 x -4 = -36 and 9 + -4 = 5. Thus:
    -6t^2 + 5t + 6 = -6t^2 + 9t - 4t + 6 = ... = (-3t - 2)(2t + 3) = -(3t + 2)(2t + 3)

    Similarly:
    B) 6t^2 + 5t + 6 cannot be factored since there is no pair of factors of 6 x 6 = 36 that add to 5.

    C) 6t^2 - 5t + 6 cannot be factored since there is no pair of factors of 6 x 6 = 36 that add to -5.

    D) -6t^2 - 5t + 6 = (-3t + 2)(2t + 3)
    (-6 x 6 = -36. -9 x 4 = -36 and -9 + 4 = -5)

    -Dan
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by zagsfan20 View Post
    5.)

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

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

    The 4th degree factor is called a "biquadratic" since it has only 4th, 2nd, and 0 degree terms in it. We can think of these as being very similar to quadratic polynomials. Let's just consider this term for a moment.

    x^4 - 10x^2 + 9

    Make the substitution y = x^2. Thus:
    x^4 - 10x^2 + 9 = y^2 - 10y + 9

    And now we can factor this like we do any other quadratic.
    y^2 - 10y + 9 = y^2 - y - 9y + 9 = ... = (y - 9)(y - 1)
    (using the same method as in the last two posts.)

    Now recall that y = x^2. So:
    x^4 - 10x^2 + 9 = y^2 - 10y + 9 = (y - 9)(y - 1) = (x^2 - 9)(x^2 - 1)

    So:
    3x^5 - 30x^3 + 27x = 3x(x^4 - 10x^2 + 9) = 3x(x^2 - 9)(x^2 - 1)

    We aren't quite done yet. Recall that a^2 - b^2 = (a + b)(a - b). The last two terms are of this form. So:
    3x^5 - 30x^3 + 27x = 3x(x^4 - 10x^2 + 9) = 3x(x^2 - 9)(x^2 - 1)  = 3x(x + 3)(x - 3)(x + 1)(x - 1)

    Now we wish to explore the zeros of this. That's simply the solution to
    3x^5 - 30x^3 + 27x = 0. Using the factored form, that's just simply all the possible factors set equal to zero:
    3x = 0 or x + 3 = 0 or x - 3 = 0 or x + 1 = 0 or x - 1 = 0.

    Thus the zeros of the polynomial are 0, \pm 3, \pm 1.

    -Dan
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  7. #7
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    Thanks a ton for the help!

    I think I'm starting to get a better understanding, but its still a work in progress.

    This is a great website, very helpful!
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