Finding the other factors of the polynomial and $K$

Question

If $(2x-1)$ is a factor of the polynomial $$p(x)=2x^3-5x^2-kx+3,$$ find $k$ and the other two factors of $p(x)$.

Answer 1

Note that $2\left(x- \frac 12\right)$ is a factor of $p(x)$, therefore we must have that $p\left(\frac 12 \right) = 0$. Now from this we get that:

$$\frac{2}{8} - \frac{5}{4} - \frac{k}{2} + 3 = 0 \implies k = 4$$

Now just divide by $(2x-1)$ to get a quadratic equation which you can solve easily. In fact by Rational Root Thorem we can see that $3$ and $-1$ are the other two zeroes of $p(x)$

Answer 2

by using the long division $$\frac{2x^3-5x^2-kx+3}{2x-1}=x^2-2x-\frac{k+2}{2}+\frac{3-\frac{k+2}{2}}{2x-1}$$ the remainder or last term must be zero or $$3-\frac{k+2}{2}=0$$ hence $$k=4$$ the other factors $$x^2-2x-3=(x-3)(x+1)$$