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Thread: Factor P(X) = 2X^5 + 6X^4 - 42X^3 - 86X^2 + 192X + 360 into linear factors

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    Unhappy Factor P(X) = 2X^5 + 6X^4 - 42X^3 - 86X^2 + 192X + 360 into linear factors

    P(X) = 2X^5 + 6X^4 - 42X^3 - 86X^2 + 192X + 360

    Use synthetic division to show how to methodically obtain the factors. Must show logical thought process.

    This problem is REALLY stumping me. Trial and error by rational roots plugins would be too extensive, grouping is just a total cluster ****, and all the GCFs (wherever they may be) are huge and/or numerous.

    It's part of a pre calc challenge. I took high school algebra about five years ago, so I don't quite remember it at an upper level. All of the videos on YouTube are VERY simplistic, and don't go into depth or use long problems such as this. I already KNOW the zeros, but just knowing the answer doesn't really help. Can someone please guide me through this?
    Last edited by mmm4444bot; 11-02-2017 at 07:38 PM. Reason: Moved function definition from title into post

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    Elite Member mmm4444bot's Avatar
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    Quote Originally Posted by breed96 View Post
    all the GCFs
    The coefficients have only one GCF. Did you begin by factoring it out of the polynomial?
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

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    Elite Member mmm4444bot's Avatar
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    Also, do you remember:

    the Rational Roots theorem?

    how to do synthetic division?
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

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    Quote Originally Posted by breed96 View Post
    P(X) = 2X^5 + 6X^4 - 42X^3 - 86X^2 + 192X + 360

    ...how to methodically obtain the factors. ...logical thought process.
    This is a little disingenuous. It should say also, "tedious" or "laborious".

    You don't seem to be recognizing "Synthetic Division". It's really not as bad as all that, especially if you know the roots already. When performing a synthetic division, those numbers along the bottom are not just window dressing. Those are your next coefficients. Don't start over with each root.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

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    Quote Originally Posted by mmm4444bot View Post
    The coefficients have only one GCF. Did you begin by factoring it out of the polynomial?
    yes. the GCF is 2

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    Quote Originally Posted by mmm4444bot View Post
    Also, do you remember:

    the Rational Roots theorem?

    how to do synthetic division?
    Isn't radical roots theorem a method of plugging in certain factors until you get zeros? And synthetic division is the drop and drag method. I know them, but my mind draws a blank when incorporating them into this equation without guessing

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    Elite Member mmm4444bot's Avatar
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    Quote Originally Posted by breed96 View Post
    yes.
    Good. So the polynomial that you're factoring is not the one posted.

    You now have:

    f(x) = 2g(x)

    You need to factor g(x).
    Last edited by mmm4444bot; 11-03-2017 at 08:20 PM. Reason: Smiley goof
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

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    Elite Member mmm4444bot's Avatar
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    Quote Originally Posted by breed96 View Post
    Isn't radical roots theorem a method of plugging in certain factors until you get zeros?
    Not quite.

    First, it's called the Rational Roots Theorem (not radical).

    Second, it's not a method of testing numbers. It's a theorem that provides you with a complete list of numbers to try (i.e., if none of the numbers in this set work, then the polynomial does not factor nicely).


    And synthetic division is the drop and drag method.
    I can see how "drag and drop method" could be used as another name for synthetic division, but I was asking whether you recall the process. My suggestion is to google keywords synthetic division examples. Refresh your memory.

    Synthetic Division is a shortcut for Polynomial Long Division (i.e., dividing one polynomial by another, to find their quotient and remainder).

    You ought to search for examples of the Rational Roots Theorem, too, but it basically uses the coefficients at each end of the polynomial to form a set of Rational numbers and then states that any Rational roots of the polynomial must appear in that set.

    Applying the theorem to g(x) yields a set of possible roots that begins like this:

    {1, -1, 2, -2, }

    Now, if 1 is a root of polynomial g, then (x-1) is a factor of g(x). If -1 is a root, then (x+1) is a factor.

    And, if (x-1) is a factor of g(x), then dividing g(x) by (x-1) yields a remainder of zero.

    So, to test whether or not 1 is a root, we divide g(x) by (x-1) using synthetic division and check to see whether the remainder is zero.

    Study some lessons and examples on-line. Let us know, if you see something you don't understand. Once you think you remember enough to try your exercise again, show us your efforts. We can go from there.

    If you'd rather review everything together, then google keywords how to use rational root theorem and synthetic division to factor polynomials. Cheers
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

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    I figured it out. I think the commentor that commented about how it would be laborious, was the one that led me to get it done. I was looking for an easy way out for a long equation like this, but I just made a couple plugin charts and did some synthetic division. Thank you!
    Last edited by stapel; 11-10-2017 at 06:29 PM.

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