Solve 1-A_2^2A_2^2/A_4^4

$$1 - \frac { A _ { 2 } ^ { 2 } A _ { 2 } ^ { 2 } } { A _ { 4 } ^ { 4 } }$$

Evaluate

$$1-A_{2}^{4}$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.

$$1-\frac{A_{2}^{4}A_{4}^{4}}{A_{4}^{4}}$$

Cancel out A_{4}^{4} in both numerator and denominator.

$$1-A_{2}^{4}$$

Factor

$$\left(A_{2}-1\right)\left(A_{2}+1\right)\left(-A_{2}^{2}-1\right)$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.

$$factor(1-\frac{A_{2}^{4}A_{4}^{4}}{A_{4}^{4}})$$

Cancel out A_{4}^{4} in both numerator and denominator.

$$factor(1-A_{2}^{4})$$

Rewrite 1-A_{2}^{4} as 1^{2}-\left(-A_{2}^{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).

$$\left(1+A_{2}^{2}\right)\left(1-A_{2}^{2}\right)$$

Reorder the terms.

$$\left(A_{2}^{2}+1\right)\left(-A_{2}^{2}+1\right)$$

Consider -A_{2}^{2}+1. Rewrite -A_{2}^{2}+1 as 1^{2}-A_{2}^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(1-A_{2}\right)\left(1+A_{2}\right)$$

Reorder the terms.

$$\left(-A_{2}+1\right)\left(A_{2}+1\right)$$

Rewrite the complete factored expression. Polynomial A_{2}^{2}+1 is not factored since it does not have any rational roots.

$$\left(-A_{2}+1\right)\left(A_{2}+1\right)\left(A_{2}^{2}+1\right)$$