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Factorize $x^4+16x-12$ over reals.

The factor is $x^4+16x-12=(x^2-2x+6)(x^2+2x-2)$

It can be factorized again but I am stuck in this step.If we want to add and then subtract we have a lot of thing to add and subtract.Another idea that I saw in books is writing as this:

$x^4+16x-12=(x^2+ax+b)(x^2+a'x+b')$

and then find $a,b,b',a'$ but there are two problems I can't find these here and we can say maybe it can factorized into one degree $3$ and one degree $1$ polynomial.

Isn't there a nice way to factor this?

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  • $\begingroup$ No, probably not, you just got a lucky quartic. $\endgroup$ – Simply Beautiful Art Dec 25 '16 at 20:02
  • $\begingroup$ @SimpleArt But it is the problem of our national math olympiad it should have a way to factor easily. $\endgroup$ – Taha Akbari Dec 25 '16 at 20:03
  • $\begingroup$ The "other idea" you mention is indeed an easy way to factor it. Why are you not satisfied with it? $\endgroup$ – Dietrich Burde Dec 25 '16 at 20:06
  • $\begingroup$ @DietrichBurde I can't find $a,b,a',b'$ $\endgroup$ – Taha Akbari Dec 25 '16 at 20:07
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    $\begingroup$ Oh, that's easy. $a=2,b=-2$ and $a'=-2,b'=6$ follow directly from the equations by comparison at $1,x,x^2$. $\endgroup$ – Dietrich Burde Dec 25 '16 at 20:08
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$$x^4+16x-12=(x^4+4x^2+4)-(4x^2-16x+16)$$

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    $\begingroup$ Well, I mean, this is easy to spot if you already know the answer. $\endgroup$ – Simply Beautiful Art Dec 25 '16 at 20:06
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    $\begingroup$ @SimpleArt, agree. I was also thinking the same thing to post as answer. But I don't think someone can strike it finely if he/she does not know the answer. $\endgroup$ – I am Back Dec 25 '16 at 20:08
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    $\begingroup$ But it is hard to discover what should we add and subtract. $\endgroup$ – Taha Akbari Dec 25 '16 at 20:08
  • $\begingroup$ you got it @TahaAkbari $\endgroup$ – I am Back Dec 25 '16 at 20:09
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    $\begingroup$ @SimpleArt However, it is also a standard procedure with the factorisation of a quartic to look for an expression as the difference of two squares, one quadratic and one linear. That is one way in which one can arrive at a cubic, and obtain the general solution of the quartic. $\endgroup$ – Mark Bennet Dec 25 '16 at 20:34
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You can check that it's not linear-times-cubic by looking for rational roots, which must have the form $p/q$ where $p$ divides 12 and $q$ divides 1. When that doesn't work, the product-of-quadratics approach is your next best bet.

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@JohnHughes explains how to find that this is not a cubic times linear factor. Once you figure that out, we know it's a quadratic times quadratic. To solve this, I'm going to go off the equation you had: $$x^4+16x-12=(x^2+ax+b)(x^2+a'x+b')$$ Expand the right side: $$x^4+16x-12=x^4+(a+a')x^3+(b+b'+aa')x^2+(ab'+ba')x+bb'$$ Now, we have $0x^3$ on the left side and $(a+a')x^3$ on the right side, so we have $a+a'=0 \implies a'=-a$. Furthermore, we have $-12$ as our constant on the left side and $bb'$ as our constant on the right side, so we have $bb'=-12 \implies b'=-\frac{12}b$. Susbtitute: $$x^4+16x-12=x^4+\left(b-\frac{12}{b}-a^2\right)x^2+\left(-\frac{12a}{b}-ab\right)x-12$$ We have $0x^2$ and $+16x$ on the left side, so by comparing such with the $x^2$ and $x$ coefficients on the right side, we get the following equations: $$b-\frac{12}{b}-a^2=0$$ $$-\frac{12a}{b}-ab=16$$ Multiply both sides by both equations $b$: $$b^2-12-ba^2=0$$ $$-12a-ab^2=16b$$

However, instead of guess and check, we can solve for $a$ in terms of $b$ in the second equation (I choose the second equation because it is easier): $$a=\frac{-16b}{12+b^2}$$ Now, remember that $b$ is a factor of $12$ since $bb'=-12$, so $b \in \{-12,-6,-4,-3,-2,-1,1,2,3,4,6,12\}$. Guess and check values of $b$ and solve for $a$. When you get an integer value of $a$, you know you've solved the problem, so substitute back into $a+a'=0$ to find $a'$ and $bb'=-12$ to find $b'$.

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    $\begingroup$ Though long but nice +1 $\endgroup$ – I am Back Dec 25 '16 at 20:18
  • $\begingroup$ @amWhy Thanks for the catch! I have edited my answer above. $\endgroup$ – Noble Mushtak Dec 25 '16 at 21:10
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What you want is to show that the resolvent cubic associated with the quartic has a rational root. Look at this method based on the usual "general solution" for quartics:

Assume a factorization if the form:

$x^4+px^2+qx+r=$

$(x^2-2\sqrt{s}x+t_1)(x^2+2\sqrt{s}x+t_2)$

Then expand the right side and match terms of like degree:

Degree 4: $1=1$ (forced)

Degree 3: $0=0$ (forced)

Degree 2: $p=-4s+(t_1+t_2)$

Degree 1: $q=(2\sqrt{s})(t_1-t_2)$

Degree 0: $r=t_1t_2$

We may treat the degree 2 and degree 1 equations as a linear system for $t_1, t_2$ in terms of $s$, then shbstitute the solution of this system into the degree 0 equation. Now that equation contains only $s$ as unknown, and after the usual simplifications we get this resolvent cubic:

$s^3-(p/2)s^2+((p^2-4r)/16)s-(q^2/64)=0$

This cubic equation has three roots corresponding to three ways the quartic roots could be paired to make quadratic factors. If you can find a rational root of this resolvent cubic you get a simplified quadratic factorization.

Here, the resolvent cubic is

$s^3+3s-4=0$

with the rational root $s=1$. Then we have a factorization

$(x^2-2x+t_1)(x^2+2x+t_2)$

where $t_1,t_2$ are determined as above with the known resolvent cubic root $s=1$ put into the degree 2 and degree 1 equations. Thus the linear system for the constant coefficients gives $t_1=6$, $t_2=-2$.

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