The polynomial, $f(x) = x^{2n} + px - 4$, where n and p are real constants, has a remainder of -8 when divided by $(x-1)$ and a remainder of 172 when divided by $(x+4)$. Find the values of n and p.

I managed to solve for p, but got stuck when finding n.

By remainder theorem, When $f(x)$ is divided by $(x-c)$, the remainder is equal to $f(c)$. $∴ f(1) = -8$

$1^{2n} + p - 4 = -8$

$p = -5$

$f(-4) = 172$

$(-4)^{2n} + (-5)(-4) -4 = 172$

$(-4)^{2n} = 156$

And I'm stuck here due to it is impossible to log a negative number.

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    $\begingroup$ $(-4)^{2n}=(-1)^{2n}(4)^{2n}$ $\endgroup$ – sharding4 May 24 '17 at 0:58
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    $\begingroup$ Are you sure the remainder isn't 272? The answer would be much nicer.. $\endgroup$ – John Doe May 24 '17 at 1:04

You have done everything right, just go on like this:

$$(-4)^{2n} = 156\to (16)^{n} = 156\to n=\log_{16}156$$

If you want $n \in \Bbb N$ the remainder would be $272$ instead of $172$. In that case we would have:

$$(16)^n=256=16^2\to n=2.$$


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