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I have figured out there is two roots between $0$ and $1 ,-1$ and $-2$ for $x^4 +3x -2 = 0$.

Therefore there should be two factors $(x + a)$ and $(y - b)$ where $a,b \in R^+$. But how to find these $a$ and $b$?

When they found I can find the next factor in $ax^2+bx+c$ form and can check for further factors easily.

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Try some general method like Descartes method or Ferrari(Check wikipedia or a google search if you are unfamiliar) – fermesomme Apr 9 '13 at 16:47
Wolfram alpha tells there are two real and two complex… – UrošSlovenija Apr 9 '13 at 16:48
up vote 7 down vote accepted


If it factors, you know the form will $(x^2 + bx \pm 1)(x^2 + cx \mp 2)$. You need a sum of $3$ and need for the cubic term to cancel out..

Now, can you use that and figure out the factors and find $b$ and $c$?

Result: $(x^2-x+2) (x^2+x-1)$

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Actually he doesn't know that the form will be that... He might check the case when the quadratics are over $\mathbb Z$, to see if it works, but he doesn't know that that is the case.. He only knows that the roots are real... – N. S. Apr 9 '13 at 23:58
Very well put! +1 – amWhy Apr 10 '13 at 0:24
@amWhy: Thank you! – Amzoti Apr 10 '13 at 0:26

For it to have some "nice" linear factors, the roots must be one of $\pm 1,\pm 2$ (this is due to the rational root theorem). You can quickly check that these are not the roots. The next bet is quadratic factors, i.e., $$(x^4+3x-2) = (x^2+ax+b)(x^2+cx+d)$$ Expanding the right hand side gives us \begin{align} a+c & = 0\\ b + d + ac & = 0\\ ad+bc & = 3\\ bd & = - 2 \end{align} This gives us the factors to be $$(x^2-x+2) \text{ and } (x^2+x-1)$$

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Over $\rm\,\Bbb Q\!:\:$ Rational Root Test excludes linear factors. Testing for quadratic factors is easy since the constant term is prime, which greatly constrains possible factors. We prove a more general case. Yours is $\rm\: p = 2,\:$ so $\rm\:3 = b = -a(p\!+\!1) = -3a\:\Rightarrow\:a = -1,\:$ so $\rm\:2 = p = 1\!+\!sa^2 = 1\!+\!s,\:$ so $\rm\:s = 1.$

Lemma $\ $ If $\rm\ f(x) = x^4 + b x-p\ $ has a quadratic factor $\rm\,\in \Bbb Q[x],\:$ and $\rm\:p\:$ is prime then

$$\rm f(x)\, =\, x^4\! -\! a(p\!+\!1)\, x\! -\! p\, =\, (s\,x^2\! +\! a\,x\!+\!p)(\color{#c00}s\,x^2\!-\!a\,x\!\color{#0a0}{-\!1}),\quad p = 1\!+\!sa^2,\ \ s =\pm1 $$

Proof $\ $ Invoking Gauss's Lemma, we may assume that it splits into monic quadratics $\rm\,\in\Bbb Z[x].\:$ Since $\rm\:p\:$ is prime, one of the factors has constant term $\rm\:\pm p,\:$ so scaling it by $\rm\,\pm1\,$ it will have the form $\rm\:s\,x^2\! +\! a\,x\!+\!p,\,\ s=\pm1 \:$. Comparing coef's, its cofactor must have leading coef $\rm\,= \color{#c00}s,\:$ constant coef $= \color{#0a0}{-1},\:$ and linear coef $\rm\ a' = - a,\:$ since the coef of $\rm\:x^3\:$ in the product $\rm\, = (a+a')s = 0.$ Finally, multiplying the two factors and comparing coef's yields the result. $\ \ $ QED

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But $f(x) = x^4 + b x-p$ can always be factored as a product of two quadratics... Your Lemma only works if you factor it over $\mathbb Z$.... – N. S. Apr 9 '13 at 23:33
@N.S. But that is the OP's problem (as I read it). There is no indication that the OP seeks factorizations over $\,\Bbb C$ or algebraic number fields. – Math Gems Apr 9 '13 at 23:49
Nope, he only knows that the roots are real...And he knows that his Polynomial has a quadratic with two real roots, why is this over $\mathbb Z$? – N. S. Apr 9 '13 at 23:55
The above formula does split the OP's polynomial into quadratic factors, from which he may then easily conclude. It is not clear what the source of the problem is. The above completely characterizes all rational factorizations of this form, which may be of interest to the OP (ditto for the method used). – Math Gems Apr 10 '13 at 0:01

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