Solve m^3-Bm^2+4m+12=0

$$m ^ { 3 } - B m ^ { 2 } + 4 m + 12 = 0$$

Solve for B

$$B=\frac{m^{3}+4m+12}{m^{2}}$$

$$m\neq 0$$

Steps for Solving Linear Equation

Subtract 4m from both sides. Anything subtracted from zero gives its negation.

$$m^{3}-Bm^{2}+12=-4m$$

Subtract 12 from both sides.

$$m^{3}-Bm^{2}=-4m-12$$

Subtract m^{3} from both sides.

$$-Bm^{2}=-4m-12-m^{3}$$

The equation is in standard form.

$$\left(-m^{2}\right)B=-m^{3}-4m-12$$

Divide both sides by -m^{2}.

$$\frac{\left(-m^{2}\right)B}{-m^{2}}=\frac{-m^{3}-4m-12}{-m^{2}}$$

Dividing by -m^{2} undoes the multiplication by -m^{2}.

$$B=\frac{-m^{3}-4m-12}{-m^{2}}$$

Divide -4m-12-m^{3} by -m^{2}.

$$B=m+\frac{4m+12}{m^{2}}$$