Solve (3a^2)(4a^6)-12

$$( 3 a ^ { 2 } ) ( 4 a ^ { 6 } ) - 12$$

Evaluate

$$12\left(a^{8}-1\right)$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 2 and 6 to get 8.

$$3a^{8}\times 4-12$$

Multiply 3 and 4 to get 12.

$$12a^{8}-12$$

Factor

$$12\left(a-1\right)\left(a+1\right)\left(a^{2}+1\right)\left(a^{4}+1\right)$$

Solution Steps

Factor out 12.

$$12\left(a^{2}a^{6}-1\right)$$

Consider a^{8}-1. Rewrite a^{8}-1 as \left(a^{4}\right)^{2}-1^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(a^{4}-1\right)\left(a^{4}+1\right)$$

Consider a^{4}-1. Rewrite a^{4}-1 as \left(a^{2}\right)^{2}-1^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

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

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

$$\left(a-1\right)\left(a+1\right)$$

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: a^{2}+1,a^{4}+1.

$$12\left(a-1\right)\left(a+1\right)\left(a^{2}+1\right)\left(a^{4}+1\right)$$