Solve 16-w^4

$$16 - w ^ { 4 }$$

Factor

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

Solution Steps

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

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

Reorder the terms.

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

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

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

Reorder the terms.

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

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

$$16-w^{4}$$