The polynomial
\[n^2 x^{2n+3} - 25n x^{n+1} + 150x^7\]
has \(x^2 - 1\) as a factor
for no values of \(n\),
for \(n=10\) only,
for \(n=15\) only,
for \(n=10\) and \(n=15\) only.
We can factorise \(x^2 -1\) as
\[x^2 -1 = (x+1)(x-1),\]
so for the polynomial
\[n^2 x^{2n+3} - 25n x^{n+1} + 150x^7\]
to have \(x^2-1\) as a factor, the value of the polynomial when \(x=\pm1\) must be \(0\) (from the factor theorem).
When \(x=+1\), we need
\[n^2 - 25n + 150=0,\]
which can be factorised as
\[(n-10)(n-15)=0,\]
so \(n=10\) or \(n=15\).
When \(x=-1\), we need
\[0 = \begin{cases} -n^2 + 25n - 150 & n \textrm{ even} \\ -n^2 - 25n - 150 & n \textrm{ odd}, \\ \end{cases}\]
which can be factorised as
\[0 = \begin{cases} -(n-10)(n-15) & n \textrm{ even} \\ -(n+10)(n+15) & n \textrm{ odd}, \\ \end{cases}\]
which gives
\[\begin{cases} n=10 & n \textrm{ even } [n \neq 15 \text{ because } n \text{ must be even]} \\ n=-15 & n \textrm{ odd } [n \neq -10 \text{ because } n \text{ must be odd]}. \\ \end{cases}\]
So for both \(x=+1\) and \(x=-1\) to be solutions of the polynomial, we must have \(n=10\).
So the answer is (b).
Alternatively we could say, when \(x = -1\) we have \(n^2+25n(-1)^{n+1} +150 =0\).
Subtracting this from \(n^2 - 25n + 150=0,\) we get \(25n((-1)^{n+1}+1)=0\), and so \(n\) must be even.
So \(n\) must be \(10\) or \(15\), and also even, so the only possibility is \(n = 10\). The answer is (b).
