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).