The value of m, in order that $x^2-\mathit{mx}-2$ is the quotient when $x^3+3x^2-4$ is divided by $x+2$ , is

We must here determine what $m$ is worth it. Divide $x^3+3x^2-4 \mathit{by} x+2$ using the long division method to find the quotient. 

The value of $m$ will then be determined by comparing the coefficients of the two quotients.
Given $\frac{x^3+3x^2-4}{x+2}= x^2-mx-2$​ 
Now $x^3$+3$x^2$−4 is divisible by $x+2$
So $x+2$ will be its one factor.

$\mathrm{LHS} = \frac{{\mathrm{x}}^3+3{\mathrm{x}}^2-4}{\mathrm{x}+2}$

$\mathrm{LHS} = \frac{{\mathrm{x}}^3+2{\mathrm{x}}^2+{\mathrm{x}}^2+2\mathrm{x}-2\mathrm{x}-4}{\mathrm{x}+2}$

$\mathrm{LHS} = \frac{{\mathrm{x}}^2(\mathrm{x}+2)+x(\mathrm{x}+2)-2x(\mathrm{x}+2)}{\mathrm{x}+2}$

$\mathrm{LHS} = \frac{(\mathrm{x} + 2)({\mathrm{x}}^2+\mathrm{x}-2)}{\mathrm{x}+2}$

$\mathrm{LHS} = ({\mathrm{x}}^2+\mathrm{x}-2)$

$R\mathrm{HS} = ({\mathrm{x}}^2-\mathrm{mx}-2)$

∴ Comparing the coefficients,

$\mathrm{m}= -1.$