Solve a^2-4b^2-2a+1/a^2-2ab-a:2a-2+4b/6a

$$\frac { a ^ { 2 } - 4 b ^ { 2 } - 2 a + 1 } { a ^ { 2 } - 2 a b - a } : \frac { 2 a - 2 + 4 b } { 6 a }$$

Evaluate

$$3$$

Short Solution Steps

Divide \frac{a^{2}-4b^{2}-2a+1}{a^{2}-2ab-a} by \frac{2a-2+4b}{6a} by multiplying \frac{a^{2}-4b^{2}-2a+1}{a^{2}-2ab-a} by the reciprocal of \frac{2a-2+4b}{6a}.

$$\frac{\left(a^{2}-4b^{2}-2a+1\right)\times 6a}{\left(a^{2}-2ab-a\right)\left(2a-2+4b\right)}$$

Factor the expressions that are not already factored.

$$\frac{6a\left(a-2b-1\right)\left(a+2b-1\right)}{2a\left(a-2b-1\right)\left(a+2b-1\right)}$$

Cancel out 2a\left(a-2b-1\right)\left(a+2b-1\right) in both numerator and denominator.

$$3$$

Factor

$$3$$