Solve w^4-16

$$w ^ { 4 } - 16$$

Factor

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

Solution Steps

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

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

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

$$\left(w-2\right)\left(w+2\right)$$

Rewrite the complete factored expression. Polynomial w^{2}+4 is not factored since it does not have any rational roots.

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

Evaluate

$$w^{4}-16$$