factoring 4th degree polynomials

• Feb 8th 2011, 12:07 AM
skoker
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.


\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.


\begin{aligned}
2n^4+14n^3+35n^2+36n+12&=(n^2+4n+4)(2n^2+6n+3) \\
\end{aligned}
• Feb 8th 2011, 12:38 AM
Unknown008
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)
• Feb 8th 2011, 01:02 AM
mr fantastic
Quote:

Originally Posted by skoker
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.


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

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 $2n^2 + 6n + 3$.
• Feb 8th 2011, 02:29 AM
skoker
@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?


\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}
• Feb 8th 2011, 03:04 AM
HallsofIvy
Quote:

Originally Posted by skoker
@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?

Quote:

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?

Quote:


\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.
$6(-2)^2+ 18(-2)+ 10= 24- 36+ 10= -2$, 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"

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

Quote:

\begin{aligned}\text{discard extra like factors.}&=(3n^2+9n+5)(2n+3)(n+2)
\end{aligned}

• Feb 8th 2011, 03:56 AM
Unknown008
The actual problem is:

Quote:

\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!
\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}
• Feb 8th 2011, 11:32 AM
skoker
@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...


\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!