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Math Help - 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.

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

    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.

    <br />
\begin{aligned}<br />
2n^4+14n^3+35n^2+36n+12&=(n^2+4n+4)(2n^2+6n+3) \\<br />
\end{aligned}<br />
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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.

    <br />
\begin{aligned}<br />
2n^4+14n^3+35n^2+36n+12&=(n^2+4n+4)(2n^2+6n+3) \\<br />
\end{aligned}<br />
    Let 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 2n^2 + 6n + 3.
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  4. #4
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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 (n+2)^2 as in perfect square trinomials?

    <br />
\begin{aligned}<br />
6n^4+39n^3+91n^2+89n+30&=(6n^3+27n^2+37n+15)(n+2) \\<br />
\text{divides evenly by -3/2.}&=(6n^2+18n+10)(2n+3)(n+2) \\<br />
\text{factor out 2 again.}&=(3n^2+9n+5)(2n+3)(n+2)(n+2) \\<br />
\text{discard extra like factors.}&=(3n^2+9n+5)(2n+3)(n+2)<br />
\end{aligned}<br />
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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 a_nz^n+ a_{n-1}z^{n-1}+\cdot\cdot\cdot+ a_1z+ a_0= 0 is of the form z= \frac{m}{n} where m evenly divides a_0 and n evenly divides a_n.

    For 2n^4+14n^3+35n^2+36n+12, that gives as possible zeros \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, and \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 (n+2)^2 as in perfect square trinomials?
    You never discard "like factors"! You never discard any factors.

    <br />
\begin{aligned}<br />
6n^4+39n^3+91n^2+89n+30&=(6n^3+27n^2+37n+15)(n+2) \\<br />
\text{divides evenly by -3/2.}&=(6n^2+18n+10)(2n+3)(n+2) \\<br />
\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.
    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 6n^2- 18n+ 10= 2(3n^2- 9n+ 5) The factor taken out is "2", NOT "n+2"

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

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

    \begin{aligned}<br />
6n^4+39n^3+91n^2+89n+30&=(6n^3+27n^2+37n+15)(n+2) \\<br />
&=(6n^2+18n+10)(2n+3)(n+2) \\\end{aligned}
    WRONG!
    \begin{aligned}<br />
6n^4+39n^3+91n^2+89n+30&=(6n^3+27n^2+37n+15)(n+2) \\<br />
&=(3n^2+9n+5)(2n+3)(n+2) \\<br />
\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...

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

    thanks for all the help!
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