Review question

# When is $n^2 x^{2n+3} - 25n x^{n+1} + 150x^7$ divisible by $x^2-1$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7036

## Solution

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 $n$,

2. for $n=10$ only,

3. for $n=15$ only,

4. 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 $n=10$.

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).