Solve 4a^2-9b^2-2a+1/4

$$4 a ^ { 2 } - 9 b ^ { 2 } - 2 a + \frac { 1 } { 4 }$$

Factor

$$\frac{\left(1-6b-4a\right)\left(1+6b-4a\right)}{4}$$

Solution Steps

Factor out \frac{1}{4}.

$$\frac{16a^{2}-36b^{2}-8a+1}{4}$$

Consider 16a^{2}-36b^{2}-8a+1. Rewrite 16a^{2}-36b^{2}-8a+1 as \left(-1+4a\right)^{2}-\left(6b\right)^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(-1+4a-6b\right)\left(-1+4a+6b\right)$$

Reorder the terms.

$$\left(4a-6b-1\right)\left(4a+6b-1\right)$$

Rewrite the complete factored expression.

$$\frac{\left(4a-6b-1\right)\left(4a+6b-1\right)}{4}$$

Evaluate

$$\frac{\left(4a-1\right)^{2}}{4}-9b^{2}$$