When is $n^2 x^{2n+3} - 25n x^{n+1} + 150x^7$ divisible by $x^2-1$?

When is $n^2 x^{2n+3} - 25n x^{n+1} + 150x^7$ divisible by $x^2-1$?

Question

The polynomial

$$n^2 x^{2n+3} - 25n x^{n+1} + 150x^7$$

has $x^2 - 1$ as a factor

1. for no values of $n$,

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1. for $n=10$ only,

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1. for $n=15$ only,

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1. for $n=10$ and $n=15$ only.

Suggestion

Given that $x^2-1$ is a factor, what linear factors of the polynomial do we know?

Can we now use the factor theorem on the polynomial twice?

We may need to be more careful with one factor than the other…

Solution

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).