# How do you find k such that f(x) = x^4 - kx^3 + kx^2 + 1 has the factor x + 2?

Apr 14, 2016

There is no Real value for $k$ such that $x + 2$ is a factor of $f \left(x\right) = {x}^{4} - k {x}^{3} + k {x}^{2} + 1$

#### Explanation:

If $\left(x + 2\right)$ is a factor of $f \left(x\right) = {x}^{4} - k {x}^{3} + k {x}^{2} + 1$
then
$\textcolor{w h i t e}{\text{XXX}} f \left(x\right) = \left(x + 2\right) \left(g \left(x\right)\right)$ for some polynomial $g \left(x\right)$
from which it is clear that if $x = \left(- 2\right)$ then $f \left(x\right) = 0$

Substituting $\left(- 2\right)$ for $x$ and noting that the result is equal to zero:
$\textcolor{w h i t e}{\text{XXX}} {\left(- 2\right)}^{2} - {\left(- 2 k\right)}^{3} + {\left(- 2 k\right)}^{2} + 1 = 0$

$\textcolor{w h i t e}{\text{XXX}} 16 {k}^{4} + 8 {k}^{3} + 4 {k}^{2} + 1 = 0$

At (local) minimum/maximum points the slope of
$\textcolor{w h i t e}{\text{XXX}} h \left(k\right) = 18 {k}^{4} + 8 {k}^{3} + 4 {k}^{2} + 1$ must be equal to zero.
Looking for values of $k$ such that
$\textcolor{w h i t e}{\text{XXX}} \frac{d \left(h \left(k\right)\right)}{\mathrm{dk}} = 64 {k}^{3} + 24 {k}^{2} + 8 k = 0$

$\textcolor{w h i t e}{\text{XXX}} 8 \left(k\right) \left(8 {k}^{2} + 3 k + 8\right) = 0$

$\textcolor{w h i t e}{\text{XXX}} \Rightarrow k = 0 \mathmr{and} \left(8 {k}^{2} + 3 k + 1\right) = 0$
$\textcolor{w h i t e}{\text{XXX}}$...but applying the quadratic formula we can see that
$\textcolor{w h i t e}{\text{XXXXXX}} \left({k}^{2} + 3 k + 1\right) = 0$ has no Real solutions.
So the only critical point occurs when $k = 0$
and $f \left(x\right)$ has a minimum value when $k = 0$

Since $f \left(x\right) > 0$ for all Real values of $x$ when $k = 0$
$\textcolor{w h i t e}{\text{XXX}}$(since, with $k = 0$, $f \left(x\right) = {x}^{4} + 1$)

Therefore there is no value of $k$ for which $\left(x + 2\right)$ is a factor of the given expression.