Solve 3m^4-48n^2

$$3 m ^ { 4 } - 48 n ^ { 2 }$$

Factor

$$3\left(m^{2}-4n\right)\left(m^{2}+4n\right)$$

Solution Steps

Factor out 3.

$$3\left(m^{4}-16n^{2}\right)$$

Consider m^{4}-16n^{2}. Rewrite m^{4}-16n^{2} as \left(m^{2}\right)^{2}-\left(4n\right)^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(m^{2}-4n\right)\left(m^{2}+4n\right)$$

Rewrite the complete factored expression.

$$3\left(m^{2}-4n\right)\left(m^{2}+4n\right)$$

Evaluate

$$3\left(m^{4}-16n^{2}\right)$$