Find the Perfect Square Trinomial 9x^2+2x

$9 x^{2} + 2 x$

Step 1

Factor out the $9$

$9 (x^{2} + \frac{2 x}{9})$

Step 2

To find the $c$ value $c = {(\frac{b}{2})}^{2}$ , divide the coefficient of $x$ by $2$ and square the result.

Multiply the numerator by the reciprocal of the denominator.

${(\frac{2}{9} \cdot \frac{1}{2})}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(\frac{2}{9} \cdot \frac{1}{2})}^{2}$

Rewrite the expression.

${(\frac{1}{9})}^{2}$

Simplify the expression.

Apply the product rule to $\frac{1}{9}$ .

$\frac{1^{2}}{9^{2}}$

One to any power is one.

$\frac{1}{9^{2}}$

Raise $9$ to the power of $2$ .

$\frac{1}{81}$

Step 3

Add $\frac{1}{81}$ to get the perfect square trinomial.

$9 (x^{2} + \frac{2 x}{9} + \frac{1}{81})$

Step 4

Simplify.

Apply the distributive property.

$9 x^{2} + 9 \frac{2 x}{9} + 9 (\frac{1}{81})$

Simplify.

Cancel the common factor of $9$ .

Cancel the common factor.

$9 x^{2} + 9 \frac{2 x}{9} + 9 (\frac{1}{81})$

Rewrite the expression.

$9 x^{2} + 2 x + 9 (\frac{1}{81})$

Cancel the common factor of $9$ .

Factor $9$ out of $81$ .

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

Cancel the common factor.

$9 x^{2} + 2 x + 9 \frac{1}{9 \cdot 9}$

Rewrite the expression.

$9 x^{2} + 2 x + \frac{1}{9}$