Factor 81x^2-18x+1

$81 x^{2} - 18 x + 1$

Rewrite $81 x^{2}$ as ${(9 x)}^{2}$ .

${(9 x)}^{2} - 18 x + 1$

Rewrite $1$ as $1^{2}$ .

${(9 x)}^{2} - 18 x + 1^{2}$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$18 x = 2 \cdot (9 x) \cdot 1$

Rewrite the polynomial.

${(9 x)}^{2} - 2 \cdot (9 x) \cdot 1 + 1^{2}$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 9 x$ and $b = 1$ .

${(9 x - 1)}^{2}$