Is the polynomial $x^4+10x^2+1$ reducible over $\mathbb{Z}[x]$?
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1Please search before asking a new question. This has been asked several times before: 1, 2 – Viktor Vaughn Jan 31 '19 at 16:27
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We see that our polynomial has no integer roots and has no real roots. So if it is reducible, there are integers $a$ and $b$ for which $$x^4+10x^2+1=(x^2+ax+1)(x^2+bx+1).$$ Can you end it now?
Acccumulation
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Michael Rozenberg
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@N. S If one of them is negative so the second is negative and we can multiply by $-1$ and to get, which I wrote. – Michael Rozenberg Jan 31 '19 at 15:58
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Your last example is impossible because our polynomial has no real roots. – Michael Rozenberg Jan 31 '19 at 15:59
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1I say again: Our polynomial has no real roots. You last example is not valid. See please better my post. – Michael Rozenberg Jan 31 '19 at 16:00
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It doesn't have any rational roots, by the rational roots test.
To see whether $f(x) = x^4+10x^2 + 1 = (x^2+ax+b)(x^2+cx+d)$ or not, try to solve the system of equations $a+c = 0$, $ac+b+d = 0$, $bd = 1$ over the integers.
Alternately, you could find the roots of $f(x)$ (they are square roots of the roots of $y^2+10y+1$) and check whether any of them has degree 2 over $\mathbb Q$.
Cardboard Box
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Here, the degree of my polynomial is 4 and it has no real root. But what about reducibility @Dane? – MAS Jan 31 '19 at 15:55
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Hey whenever the expression contains even powers of $x$ you can always take $x^2=t$, to reduce the equation to : $$t^2+10t+1$$ which is a quadratic polynomial over $t$
Now follow the usual factorization as you would for the quadratic expression....
$$t^2+10t+1=(t- \alpha)(t- \beta)$$ where $$\alpha =\frac{-10+\sqrt{96}}{2}$$
$$\beta=\frac{-10-\sqrt{96}}{2}$$ Now put back $t=x^2$ and now you will see that the factors will lead to two more factors
$$(x^2-\alpha)(x^2-\beta)$$ which will give you 4 factors: $$(x-\alpha)(x+\alpha)(x-\beta)(x+\beta)$$
Hope this helps …...
SNEHIL SANYAL
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