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I have this expression: $3m^4-6m^3+14m^2-6m+11=0$ and I want to factorize it in $(m^2+1)(3m^2-6m+11)$. How can I do it? Thanks for any help!

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Please clarify: are you looking to show that that is the factorization, or are you asking how you would have discovered that factorization if you hadn't already known about it? – Hurkyl Mar 25 '13 at 17:18

Hint $\ $ Because the leading and constants coefficients are primes, the possible factors are highly constrainted, so we can quickly find quadratic factors by undetermined coefficients. First, notice $\rm\:mod\ 3\!:\ -f \equiv x^2+1,\:$ so we check for a factor of form $\rm\: x^2\! +\, 3a\, x + 1.\:$ Its cofactor must have leading coefficient $3$ and constant coefficient $11$, i.e.

$$\rm\begin{eqnarray} (x^2+3a\,x+1)(3\,x^2\!+b\,x+11) &\,=\:&\rm 3\,x^4 + (b\!+\!9a)\, x^3 + (14\!+\!3ab)\, x^2 + (b\!+\!33a)\,x + 11 \\ &=&\rm 3\, x^2 - 6\, x^3 + 14\, x^2 - 6\,x + 11\end{eqnarray}$$

Comparing $\rm\:x^2$ terms, $\rm\:14=14\!+\!3ab\Rightarrow ab=0\:$ so $\rm\,a=0\,$ or $\rm\,b=0.\:$ If $\rm\:b=0\:$ then comparing $\rm\,x^3$ terms, $\rm\:9a=-6,\,$ contra $\rm\:a\in \Bbb Z.\:$ Thus $\rm\:a=0.\:$ Comparing $\rm\,x^3$ terms, $\rm\:b = -6,\:$ which works.

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Nice answer. I like your unique formatting style! – The Chaz 2.0 Mar 26 '13 at 0:45





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But this appears to pull the solution out of a hat. You should say why you tried that particular factorization. See my answer for one such explanation. – Math Gems Mar 25 '13 at 17:26

Let $f$ be the polynomial. If you know about complex numbers, you can compute $f(i) = 0$, showing you that $f$ is divisible by $(m - i)$. Since the coefficients are real, also $-i$ is a root, and so $f$ is divisible by $(m + i)$, too. Therefore $f$ is divisible by $(m - i)(m + i) = m^2 + 1$. Polynomial long division now gives you the factorization.

EDIT: How do you see $f(i) = 0$? If $f = \sum_{i=0}^d a_i X^i$ with real coefficients $a_i$, this is quite easy to check: $f(i) = 0$ if and only if the two alternating sums $$a_0 - a_2 + a_4 - a_6 \pm \ldots\quad\text{and}\quad a_1 - a_3 + a_5 - a_7 \pm \ldots$$ both equal zero. In this example, $11-14 + 3 = 0$ and $(-6)-(-6) = 0$, so $f(i) = 0$.

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I'm not sure how seeing $f(i) = 0$ is any easier than seeing that $(m^2+1)$ is a factor outright. – Andrew Salmon Mar 25 '13 at 17:20
@AndrewSalmon: Well, since the coefficients of $f$ are real, both conditions are equivalent. With this knowledge, you see $f(i) = 0$ if and only if you see $m^2 + 1 \mid f$. – azimut Mar 25 '13 at 17:23
... And the point is that many students would arguably not see either! – The Chaz 2.0 Mar 25 '13 at 18:23
@TheChaz2.0 For a polynomial $f = a_0 + a_1 X + a_2 X^2 + \ldots$ with real coefficients, the check for $f(i) = 0$ is routine and almost as easy as $f(1) = 0$ or $f(-1) = 0$: $f(i) = 0$ means that the two alternating sums $a_0 - a_2 + a_4 - a_6 \pm\ldots$ and $a_1 - a_3 + a_5 - a_7 \pm\ldots$ both equal zero. In this case, $3 - 14 + 11 = 0$ and $-6 -(-6) = 0$. – azimut Mar 25 '13 at 23:28
@TheChaz2.0: Thank you. Seems that the trick with the alternating sums is not quite well-known; I will include it in the anser. On yor question: The strategy for factoring polynomials is typically trying all your tools, from the easier to the more time-consuming ones. Recovering the factors from the product of two random irreducible cubic polynomials is quite hard. In general, one has to apply a technique like in the answer of MathGems, I guess. However, for the polynomial in this question, noticing $f(i) = 0$ is faster. – azimut Mar 26 '13 at 0:11

You already factorized the $3m^4 −6m^3 +14m^2 −6m+11$ with $(m^2 +1)(3m^2 −6m+11)$.
First factor equals to $(m^2 +1)$ and second $(3m^2 −6m+11)$. $(m^2 +1)(3m^2 −6m+11) = 3m^4-6m^3+11m^2+3m^2-6m+11 = 3m^4-6m^3+14m^2+11$

But if you need solve $3m^4 −6m^3 +14m^2 −6m+11 = 0$ using factorization $(m^2 +1)(3m^2 −6m+11)$ it's a little different problem.
$(m^2 +1)(3m^2 −6m+11) = 0$ is true if 1.:$m^2 +1 = 0$ or 2.:$3m^2 −6m+11=0$.
1. $m^2 +1 = 0$ => $m_{1,2} = \pm\sqrt{-1} = \pm{i}$
2. $3m^2 −6m+11=0$ => $m_{3,4} = 1\pm{2}i\sqrt{\frac{2}{3}}$

Here are four solutions:
1. $m = i$ : $3i^4-6i^3+14i^2-6i+11 = 3+6i+14-6i+11 = 0$
2. $m = -i$ : $3(-i)^4-6(-i)^3+14(-i)^2-6(-i)+11$
$ = 3(-1)^4i^4-6(-1)^3i^3+14(-1)^2i^2-6(-i)+13 = 3 -6i-14+6i+13 = 0$
3. $m = 1+{2}i\sqrt{\frac{2}{3}}$: substitute to equation as in 1. or 2.
4. $m = 1-{2}i\sqrt{\frac{2}{3}}$: substitute to equation as in 1. or 2.

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