Solve c^2d^2-16c^2-4d^2+64

$$c ^ { 2 } d ^ { 2 } - 16 c ^ { 2 } - 4 d ^ { 2 } + 64$$

Factor

$$\left(c-2\right)\left(d-4\right)\left(c+2\right)\left(d+4\right)$$

Solution Steps

Do the grouping c^{2}d^{2}-16c^{2}-4d^{2}+64=\left(c^{2}d^{2}-16c^{2}\right)+\left(-4d^{2}+64\right), and factor out c^{2} in the first and -4 in the second group.

$$c^{2}\left(d^{2}-16\right)-4\left(d^{2}-16\right)$$

Factor out common term d^{2}-16 by using distributive property.

$$\left(d^{2}-16\right)\left(c^{2}-4\right)$$

Consider d^{2}-16. Rewrite d^{2}-16 as d^{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).

$$\left(d-4\right)\left(d+4\right)$$

Consider c^{2}-4. Rewrite c^{2}-4 as c^{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).

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

Rewrite the complete factored expression.

$$\left(d-4\right)\left(c-2\right)\left(c+2\right)\left(d+4\right)$$

Evaluate

$$\left(c^{2}-4\right)\left(d^{2}-16\right)$$