Solve 16-d^4

$$16 - d ^ { 4 }$$

Factor

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

Solution Steps

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

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

Reorder the terms.

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

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

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

Reorder the terms.

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

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

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

Evaluate

$$16-d^{4}$$