Question

Solve: $(8x^3-5x^2+6x)÷2x$

Answer 1

$(8x^3-5x^2+6x)÷2x$
$=\frac{8x^3-5x^2+6x}{2x}$
$=\frac{8x^3}{2x}-\frac{5x^2}{2x}+\frac{6x}{2x}$
$=4x^2-\frac{5}{2}x+3$
Alternative Method
For $8x^3-5x^2+6x$, separating $2x$ from each term we get,
$8x^3-5x^2+6x=2x\left(4x^2\right)-2x\left(\frac{5}{2}x\right)+2x\left(3\right)$
$=2x(4x^2-\frac{5}{2}x+3)$
$\frac{8x^3-5x^2+6x}{2x}=\frac{2x(4x^2-\frac{5}{2}x+3)}{2x}$
$=4x^2-\frac{5}{2}x+3$