Factor by Grouping 9x^2-6x+1

$9 x^{2} - 6 x + 1$

Step 1

Rewrite $9 x^{2}$ as ${(3 x)}^{2}$ .

${(3 x)}^{2} - 6 x + 1$

Step 2

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

${(3 x)}^{2} - 6 x + 1^{2}$

Step 3

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

$6 x = 2 \cdot (3 x) \cdot 1$

Step 4

Rewrite the polynomial.

${(3 x)}^{2} - 2 \cdot (3 x) \cdot 1 + 1^{2}$

Step 5

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 3 x$ and $b = 1$ .

${(3 x - 1)}^{2}$