Question

Divide:

$\frac{2}{3}{z}^4-\frac{1}{3}{z}^2-1$ by $\frac{1}{3}z$

• A. $2z^3+z-\frac{3}{z}$
• B. $2z^3-z-\frac{3}{z}$
• C. $z^3-z-\frac{2}{z}$
• D. $z^3-z-\frac{3}{z}$

Answer 1

The correct option is B $2z^3-z-\frac{3}{z}$

$\frac{2}{3}{z}^4-\frac{1}{3}{z}^2-1÷\frac{1}{3}z$
$=\frac{\frac{2}{3}{z}^4}{\frac{1}{3}z}-\frac{\frac{1}{3}{z}^2}{\frac{1}{3}z}-\frac{1}{\frac{1}{3}z}$
$=2z^3-z-\frac{3}{z}$