2026/063) Factor $n^4+6n^3+11n^2+6n+1$

Let $f(n) =  n^4+6n^3+11n^2+6n+1$

This does not have a change of sign so there is no positive root

So it can have -ve root if it has real root it has to be -1

Checking using the rational root theorem $(f(-1) = 1 - 6 + 11 -6 +1 = 1$

So -1  is not a root

So it has to be product of 2 quadratic polynomials 

This is a quartic polynomal. And the coefficients are symmetric.

So if n is a root then $\frac{1}{n}$ is a root. So the coeffcient of$x^2$ and constant should same in both polynomials. However I shall continue as below

$f(n) =  n^4+6n^3+11n^2+6n+1$

$=  (n^4+1) +11n^2+6(n^3+n)$ reordering the terms

$= (n^2+1)^2 -2n^2 +11n^2+6n(n^2+1)$ putting $n^4+1$ in terms of $(n^2+1)$

$=(n^2+1)^2 + 6n(n^2+1) + 9n^2$

$= (n^2+1)^2+ 2(3n)(n^2+1) + (3n)^2$ geting expressing in $a^2+2ab+b^2$ form

$=(n^2+1+3n)^2$

$= (n^2+3n+1)^2$ putting in standard form