How do you find the quotient of (2h^3+8h^2-3h-12)div(h+4)?

$2 {h}^{2} - 3$

Explanation:

I'm going to rewrite this as:

$\frac{2 {h}^{3} + 8 {h}^{2} - 3 h - 12}{h + 4}$

We can factor the numerator this way:

$\frac{\left(2 {h}^{3} + 8 {h}^{2}\right) + \left(- 3 h - 12\right)}{h + 4}$

$\frac{2 {h}^{2} \left(h + 4\right) + \left(- 3\right) \left(h + 4\right)}{h + 4}$

$\frac{\left(2 {h}^{2} - 3\right) \left(h + 4\right)}{h + 4}$

We can now cancel the $\left(h + 4\right)$ on top and bottom:

$\frac{\left(2 {h}^{2} - 3\right) \cancel{\left(h + 4\right)}}{\cancel{\left(h + 4\right)}}$

And be left with:

$2 {h}^{2} - 3$