Solve 16-y^4

$$16 - y ^ { 4 }$$

Factor

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

Solution Steps

Rewrite 16-y^{4} as 4^{2}-\left(-y^{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).

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

Reorder the terms.

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

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

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

Reorder the terms.

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

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

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

Evaluate

$$16-y^{4}$$