hey i didnt really understand the term non constant

vikasvak wrote: hey i didnt really understand the term non constant degree $>0$

Possibility 1 We write the polynomial $P(x)$ as: \[P(x)=x^{5}-nx-n-2=(x-a)(x^{4}+bx^{3}+cx^{2}+dx+e)\] For $x=a$ is: \[P(a)=a^{5}-na-n-2=0\] \[a^{5}-na=n+2\] \[a(a^{4}-n)=n+2\] If $a^{4}-n=e$, then $a\cdot e=n+2$ \[ \left\{\begin{array}{l}{a^{4}-n=e}\\ {a\cdot e=n+2}\end{array}\right. \Rightarrow \left\{\begin{array}{l}{a^{4}-e=n}\\ {a\cdot e=n+2}\end{array}\right. \] Subtracting: \[a^{4}-e-a\cdot e=-2\] or\[a^{4}-a\cdot e=e-2\]and \[a(a^{3}-e)=e-2\] If $a^{3}-e=l$, then $a\cdot l=e-2$ \[ \left\{\begin{array}{l}{a^{3}-e=l}\\ {a\cdot l=e-2}\end{array}\right. \Rightarrow \left\{\begin{array}{l}{a^{3}-l=e}\\ {a\cdot l=e-2}\end{array}\right. \] Subtracting: \[a^{3}-l-a\cdot l=2\] or\[a^{3}-a\cdot l=l+2\]and \[a(a^{2}-l)=l+2\] If $a^{2}-l=t$, then $a\cdot t=l+2$ \[ \left\{\begin{array}{l}{a^{2}-l=t}\\ {a\cdot t=l+2}\end{array}\right. \Rightarrow \left\{\begin{array}{l}{a^{2}-t=l}\\ {a\cdot t=l+2}\end{array}\right. \] Subtracting: \[a^{2}-t-a\cdot t=-2\] or\[a^{2}-a\cdot t=t-2\]and \[a(a-t)=t-2\] If $a-t=z$, then $a\cdot z=t-2$ \[ \left\{\begin{array}{l}{a-t=z}\\ {a\cdot z=t-2}\end{array}\right. \Rightarrow \left\{\begin{array}{l}{a-z=t}\\ {a\cdot z=t-2}\end{array}\right. \] Subtracting: \[a-z-a\cdot z=2\] or\[a-a\cdot z=z+2\]and \[a(1-z)=z+2\] Finally \[a=\frac{z+2}{1-z}\] We know that a,n,e,l,t and z are integers The only integer solutions (z,a) are: (-2,0), (0,2), (2,-4) and (4,-2) We find for a: 2, -2, -4 and for n: 10, 34, 342 \[P(x)=x^{5}-10x-12=(x-2)(x^{4}+2x^{3}+4x^{2}+8x+6)\] \[P(x)=x^{5}-34x-36=(x+2)(x^{4}-2x^{3}+4x^{2}-8x-18)\] \[P(x)=x^{5}-342x-344=(x+4)(x^{4}-4x^{3}+16x^{2}-64x-86)\] Possibility 2 We write the polynomial $P(x)$ as: \[P(x)=x^{5}-nx-n-2=(x^{2}+ax+b)(x^{3}+cx^{2}+dx+e)\] \[P(x)=x^{5}+(a+c)x^{4}+(ac+b+d)x^{3}+(ad+bc+e)x^{2}+(ae+bd)x+be\] We substitute $c=-a$: \[P(x)=x^{5}-(a^{2}-b-d)x^{3}+(ad-ab+e)x^{2}+(ae+bd)x+be\] We substitute $d=a^{2}-b$ \[P(x)=x^{5}+(a^{3}-2ab+e)x^{2}+(a^{2}b+ae-b^{2})x+be\] At last: $e=a(2b-a^{2}$ \[P(x)=x^{5}-(a^{4}-3a^{2}b+b^{2})x-ab(a^{2}-2b)\] Because $P(x)=x^{5}-nx-n-2$, is: \[a^{4}-3a^{2}b+b^{2}=n and ab(a^{2}-2b)=n+2\] Subtracting: \[a^{4}-3a^{2}b+b^{2}-ab(a^{2}-2b)=-2\] It's a quadratic equation in b; \[(2a+1)b^{2}-(a+3)a^{2}b+a^{4}+2=0\] The only integer solution is $a=-1$, $b=-3$ and $n=19$: \[P(x)=x^{5}-19x-21=(x^{2}-x-3)(x^{3}+x^{2}+4x+7)\]

Very nice.

A fair amount of calculations can be avoided if $x$ is substituted by $x-1$ and $n$ by $5-n$. The polynomial becomes $x^5-5x^4+10x^3-10x^2+nx-3$. The 4 (not only 3) factorizations with a linear factor can be read off directly. In order to find the remaining 2 (not only 1) factorizations into a quadratic and a cubic polynomial you only have to find the integer roots of a cubic polynomial.

can we use Eisenstein Criterion here?

Assume x-c is a factor; use synthetic division p(x)/x-c since we know the remainder is zero, we can solve for possible values of c and n!