# How do you use the remainder theorem to see if the n+8 is a factor of n^4+9n^3+14n^2+50n+9?

Aug 19, 2016

$\left(n + 8\right)$ is not a factor of $f \left(n\right) = {n}^{4} + 9 {n}^{3} + 14 {n}^{2} + 50 n + 9$ since $f \left(- 8\right) \ne 0$

#### Explanation:

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$f \left(n\right) = {n}^{4} + 9 {n}^{3} + 14 {n}^{2} + 50 n + 9$

By the remainder theorem, $\left(n - a\right)$ is a factor of $f \left(n\right)$ if and only if $f \left(a\right) = 0$.

So in our case $\left(n + 8\right)$ is a factor if $f \left(- 8\right) = 0$.

Note that if $n$ is even, then all of the first four terms are even, but the constant term is odd. So $f \left(n\right)$ is odd and thus non-zero.

So $\left(n + 8\right)$ is not a factor of $f \left(n\right)$.

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Bonus

By the rational roots theorem, any rational zeros of $f \left(n\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $9$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 3 , \pm 9$

When $n = + 1$ we have $3$ odd terms and two even ones, so the sum is odd and therefore non-zero.

When $n = \pm 3$ then first three terms and the last (constant) term are all divisible by $9$, but the fourth term is not. So $f \left(n\right)$ is not divisible by $9$ and therefore non-zero.

When $n = \pm 9$ then the first three terms are all divisible by ${9}^{2}$. The sum of the last two terms is $9 \left(1 \pm 50\right)$, which is a multiple of $9$ and possibly $27$, but not $81$. So $f \left(n\right)$ is not divisible by ${9}^{2}$ and therefore non-zero.

Hence $f \left(n\right)$ has no rational zeros, so no rational linear factors.