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I would like to ask if someone could help me with the following equation. \begin{equation} x^4+ax^3+(a+b)x^2+2bx+b=0 \end{equation} Could you first solve in general then $a=11$ and $b=28$. I get it to this form but I stuck. \begin{equation} 1+a\left(\frac{1}{x}+\frac{1}{x^2}\right)+b\left(\frac{1}{x^2}+\frac{1}{x^3}+\frac{1}{x^3}+\frac{1}{x^4}\right) = 0 \end{equation} Thank you in advance.

JohnColtraneisJC
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    Why do you believe that there will be a nice solution? There does exist a generalized solution but hardly anyone ever uses or learns it beyond learning of its existence. – JMoravitz Sep 25 '19 at 18:53
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    For the special case of $a=11,b=28$ and the problem having been assigned by a teacher or a textbook, you could pray that one or more of the roots would be a small negative integer such as $-1$ or $-2$ (cough cough) noting that any real roots would have to be negative, and if you are correct then you could perform long division, reducing this to a cubic or a quadratic and continue from there with more standard approaches. – JMoravitz Sep 25 '19 at 19:00
  • $u = \left(\frac{1}{x}+\frac{1}{x^2}\right)$, so the plynomial looks like this $1+au+b\left(\frac{u}{x}+\frac{u}{x^2}\right) = 0$, but i dont know how to continue. –  Sep 25 '19 at 19:04
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    Assume $x \ne -1$, you can rewrite your equation as $$x^4 + ax^2(x+1)+b(x+1)^2 = 0 \iff \left(\frac{x^2}{x+1}\right)^2 + a\left(\frac{x^2}{x+1}\right) + b = 0$$ – achille hui Sep 25 '19 at 19:05
  • Could you explain to me? Please. –  Sep 25 '19 at 19:07
  • How do you come up with $x^4+ax^2(x+1)+b(x+1)^2$. –  Sep 25 '19 at 19:15

2 Answers2

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The polynomial equation for $(a, b)=(11,28)$ is given by $$ x^4+11x^3+39x^2+56x+28=(x^2 + 7x + 7)(x + 2)^2=0. $$ "How do we come up wit this layout"? By the rational root test, we find the factor $x-2$ twice, and dividing gives the factor $x^2+7x+7$. Hence the roots are $x=-2,-2,\frac{\sqrt{21}-7}{2},\frac{-\sqrt{21}-7}{2}$.

In general, the polynomial will not have any integral roots. A good example is the case $(a, b)=(1,1)$, where the polynomial $$ x^4+x^3+2x^2+2x+1 $$ is irreducible over $\Bbb Q$. It has no real root at all.

Dietrich Burde
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You are almost there !

$$1+a\left(\frac{1}{x}+\frac{1}{x^2}\right)+b\left(\frac{1}{x^2}+\frac{1}{x^3}+\frac{1}{x^3}+\frac{1}{x^4}\right) = 0$$ $$1+a\left(\frac{x+1}{x^2}\right)+b\left(\frac{x+1}{x^3}+\frac{x+1}{x^4}\right) = 0$$ $$1+a\left(\frac{x+1}{x^2}\right)+b\left(\frac{x+1}{x^2}\right)^2 = 0$$

albert chan
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    I am not sure it is obvious how did get the first equation. – NoChance Sep 25 '19 at 22:14
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    This is from OP. x^4 + a(x^3+x^2) + b(x^2 + 2x + 1) = 0. Divide it by x^4. – albert chan Sep 25 '19 at 22:37
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    Thanks. How to go on to find $x$? – NoChance Sep 25 '19 at 23:36
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    After above quadratic solved, we have 2 roots, say r1, r2. Solved 2 more quadratics, r1x^2 = x+1, r2x^2 = x+1, and we have all 4 roots. Note: r1, r2 might be complex. It is harder to solve, but you only need to solve 1, and doubled the roots with complex conjugates. – albert chan Sep 26 '19 at 00:21
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    Thank you for the explanation but the last equation would be transformed to a degree 4 equation, which is not easier to solve than the original. – NoChance Sep 26 '19 at 06:19