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Thread: factoring 4th degree polynomials

  1. #1
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    factoring 4th degree polynomials

    I am trying to factor 2 4th degree polynomials. I am stuck on both of them for different reasons.

    1) I can use synthetic division and rational zero factor -3/2 to get to 2th degree. but that does not give the right quotient. multiplying the factor back it gives a different polynomial.

    $\displaystyle
    \begin{aligned}
    6n^4+39n^3+91n^2+89n+30&=(6n^3+27n^2+37n+15)(n+2) \\
    \text{divides evenly by -3/2}&=(6n^2+18n+10)(2n+3)(n+2) \\
    \text{multiply factors back}&=(wrong!)(n+2)
    \end{aligned}
    $

    2) this ones is from a book, it shows 2 solution steps from 4th degree to 2 quadratic equations? how do you do that? I tried all kinds of factoring and can't figure it out.

    $\displaystyle
    \begin{aligned}
    2n^4+14n^3+35n^2+36n+12&=(n^2+4n+4)(2n^2+6n+3) \\
    \end{aligned}
    $
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  2. #2
    MHF Contributor Unknown008's Avatar
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    1) The quotient should be (3n^2 + 9n + 5)

    You didn't divide by 2

    2) I don't know either, but from what I see, n^2 + 4n + 4 is a perfect square (ie you get 144 which is a square number when n = 1)

    Which suggests that they tried in (-2) and found that it worked for the polynomial once more after factoring out (x+2)
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  3. #3
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    mr fantastic's Avatar
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    Quote Originally Posted by skoker View Post
    I am trying to factor 2 4th degree polynomials. I am stuck on both of them for different reasons.

    [snip]

    2) this ones is from a book, it shows 2 solution steps from 4th degree to 2 quadratic equations? how do you do that? I tried all kinds of factoring and can't figure it out.

    $\displaystyle
    \begin{aligned}
    2n^4+14n^3+35n^2+36n+12&=(n^2+4n+4)(2n^2+6n+3) \\
    \end{aligned}
    $
    Let $\displaystyle f(n) = 2n^4+14n^3+35n^2+36n+12$. Note that f(-2) = 0 therefore (n + 2) is a factor. Use polynomial long division to get the cubic factor g(n).

    It turns out that g(-2) = 0 therefore (n + 2) is a factor of g(n). Use polynomial long division to get the quadratic factor of g(n). This factor will turn out to be $\displaystyle 2n^2 + 6n + 3$.
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    @mr fantastic
    so to find f(-2) = 0 did you just guess based on leading coefficient and constant factors? to do it without knowing what the factors were...

    here is what I get now. factor out 2 'again' so there are 2 factors of (n+2) and a total of 4 linear factors. because (n+2)(n+2) are like factors you only keep one? so to expand the polynomial again discard duplicate factors and multiply.

    how do you know when do discard like factors and when to $\displaystyle (n+2)^2$ as in perfect square trinomials?

    $\displaystyle
    \begin{aligned}
    6n^4+39n^3+91n^2+89n+30&=(6n^3+27n^2+37n+15)(n+2) \\
    \text{divides evenly by -3/2.}&=(6n^2+18n+10)(2n+3)(n+2) \\
    \text{factor out 2 again.}&=(3n^2+9n+5)(2n+3)(n+2)(n+2) \\
    \text{discard extra like factors.}&=(3n^2+9n+5)(2n+3)(n+2)
    \end{aligned}
    $
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  5. #5
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    Quote Originally Posted by skoker View Post
    @mr fantastic
    so to find f(-2) = 0 did you just guess based on leading coefficient and constant factors? to do it without knowing what the factors were...
    I suspect you are referring to the "rational root theorem": "Any rational root of the polynomial equation $\displaystyle a_nz^n+ a_{n-1}z^{n-1}+\cdot\cdot\cdot+ a_1z+ a_0= 0$ is of the form $\displaystyle z= \frac{m}{n}$ where m evenly divides $\displaystyle a_0$ and n evenly divides $\displaystyle a_n$.

    For $\displaystyle 2n^4+14n^3+35n^2+36n+12$, that gives as possible zeros $\displaystyle \pm 1$, $\displaystyle \pm 2$, $\displaystyle \pm 3$, $\displaystyle \pm 4$, $\displaystyle \pm 6$, $\displaystyle \pm 12$, and $\displaystyle \pm \frac{1}{2}$. Is that what you meant by "guess based on leading coefficient and constant factors"? Not really "guessing" is it?

    here is what I get now. factor out 2 'again' so there are 2 factors of (n+2) and a total of 4 linear factors. because (n+2)(n+2) are like factors you only keep one? so to expand the polynomial again discard duplicate factors and multiply.

    how do you know when do discard like factors and when to $\displaystyle (n+2)^2$ as in perfect square trinomials?
    You never discard "like factors"! You never discard any factors.

    $\displaystyle
    \begin{aligned}
    6n^4+39n^3+91n^2+89n+30&=(6n^3+27n^2+37n+15)(n+2) \\
    \text{divides evenly by -3/2.}&=(6n^2+18n+10)(2n+3)(n+2) \\
    \text{factor out 2 again.}&=(3n^2+9n+5)(2n+3)(n+2)(n+2) \end{aligned}$
    Here's your mistake- you didn't factor out "2" you factored out "n+ 2" and that is wrong.
    $\displaystyle 6(-2)^2+ 18(-2)+ 10= 24- 36+ 10= -2$, not 0 so "n-(-2)= n+ 2" is NOT a factor.
    What is true is that $\displaystyle 6n^2- 18n+ 10= 2(3n^2- 9n+ 5)$ The factor taken out is "2", NOT "n+2"

    $\displaystyle (6n^2+ 18n+ 5)(2n+ 3)(n+2)= 2(3n^2+ 9n+ 5)(2n+3)(n+2)$

    $\displaystyle \begin{aligned}\text{discard extra like factors.}&=(3n^2+9n+5)(2n+3)(n+2)
    \end{aligned}
    $
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  6. #6
    MHF Contributor Unknown008's Avatar
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    The actual problem is:

    $\displaystyle \begin{aligned}
    6n^4+39n^3+91n^2+89n+30&=(6n^3+27n^2+37n+15)(n+2) \\
    &=(6n^2+18n+10)(2n+3)(n+2) \\\end{aligned}$
    WRONG!
    $\displaystyle \begin{aligned}
    6n^4+39n^3+91n^2+89n+30&=(6n^3+27n^2+37n+15)(n+2) \\
    &=(3n^2+9n+5)(2n+3)(n+2) \\
    \end{aligned}$
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  7. #7
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    @HallsofIvy
    yes I mean guess which factors using rational zero theorem.

    OK! so as was stated more then one time I re-factor and take out 2 and (x+2). I trick myself with all the operations and sine change etc trying to get 4 linear factors by the fundamental theorem. so now I get...

    $\displaystyle
    \begin{aligned}
    6n^4+39n^3+91n^2+89n+30&=(6n^3+27n^2+37n+15)(n+2) \\
    &=(6n^2+18n+10)\bigg( n+ \frac{3}{2} \bigg)(n+2) \\
    \text{factor out "2".}&=(3n^2+9n+5)(2)\bigg( n+ \frac{3}{2}\bigg)(n+2) \\
    \text{combine factors to clear fraction}&=(3n^2+9n+5)(2n+3)(n+2)
    \end{aligned}
    $

    thanks for all the help!
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