Factoring the quintic $n^5-16n^4+95n^3-260n^2+324n-144$

Question

I was attempting to solve $n^5-16n^4+95n^3-260n^2+324n-144=0$ but then realised I didn't know how to.

How would one go about factoring such a quintic and solve for n? I know that the factored form is $(n-1)(n-2)(n-3)(n-4)(n-6)$, but I am unaware of a good process of factoring it to get the solution. Thanks in advance.

Answer 1

In such cases I usually divide the constant term by the leading coefficient in this case we have -144/1 which is -144 and then proceed to find p(a) where a is any factor of -144. By trial and error we find p(1) is 0 and p(2) is also 0 and p(3) is also 0 (we get lucky in this case because they're all small numbers) then you can divide the original quintic equation by (x-1)(x-2)(x-3) and get a quadratic which is going to give you the rest of the factors. Hope this helps.

Answer 2

Method 1 In general factoring quintics can be time-intensive, but it's always easy to check for rational roots of a polynomial with integer coefficients: In the special case that a polynomial $p$ is monic (i.e., its leading coefficient is $1$) the Rational Root Theorem says that if $r$ is a root of $p$, then $r$ is a factor of the constant coefficient $a_0$.

Since the signs of $p(n)$ alternate, we know that all real roots of $p$ are positive, leaving $15$ possibilities $(1, 2, \ldots, 72, 144)$. Optionally, a bound of Lagrange then implies that all roots are $\leq 32$, leaving just $11$ possibilities. Trying the first possibility gives that $p(1)$ (the sum of the coefficients) is already $0$, so $(n - 1) \mid p(n)$, and polynomial long division gives $$p(n) = (n - 1) (n^4 - 15 n^3 + 80 n^2 - 180 n + 144) .$$ As Oscar Lanzi pointed out in a comment, the quartic factor has the special form $$n^4 + b n^3 + c n^2 + d n + \left(\frac{d}{b}\right)^2 ,$$ so we can look for factorizations of the form $$n^4 + b n^3 + c n^2 + d n + \left(\frac{d}{b}\right)^2 = \left(n^2 + \alpha n + \frac{d}{b}\right) \left(n^2 + (b - \alpha) n + \frac{d}{b}\right).$$ Expanding the factorization gives that $\alpha$ need only satisfy $c = \frac{2 d}{b} + \alpha b - \alpha^2$, and substituting our values of $b, c, d$ gives the solutions $\alpha = -7, -8$. So, our quartic factors as $$(n^2 - 7 n + 12)(n^2 - 8 n + 12) = (n - 3) (n - 4) (n - 2) (n - 6),$$ hence fully factoring the quintic into linear factors.

Method 2 Rewrite the polynomial in terms of a new variable, $m = n - a$ for which the coefficients are more manageable. Taking $a = 3$ gives a relatively tractable polynomial: \begin{align*} m^5 - m^4 - 7 m^3 + m^2 + 6 m &= m (m^4 - m^3 - 7 m^2 + m + 6) \\ &= m [(m^4 - m^2) - (m^3 - m) - (6 m^2 - 6)] \\ &= m [m^2 (m^2 - 1) - m (m^2 - 1) - 6 (m^2 - 1)] \\ &= m (m^2 - m - 6) (m^2 - 1) \\ &= m (m - 3) (m + 2) (m + 1) (m - 2) . \end{align*} In terms of $n = m + 3$ the factorization is (after rearranging) $$(n - 1) (n - 2) (n - 3) (n - 4) (n - 6).$$