Given the equation function $\displaystyle f(x)=4x^3-kx^2+x+3 $, find the value of $\displaystyle k$ such that $\displaystyle (2x+1)$ is a factor
If $\displaystyle 2x + 1$ is a factor, then by the factor and remainder theorems...
$\displaystyle f\left(-\frac{1}{2}\right) = 0$.
So we get
$\displaystyle 4\left(-\frac{1}{2}\right)^3 - k\left(-\frac{1}{2}\right)^2 - \frac{1}{2} + 3 = 0$
$\displaystyle -\frac{1}{2} - \frac{1}{4}k + \frac{5}{2} = 0$
$\displaystyle 2 = \frac{1}{4}k$
$\displaystyle 8 = k$.