Find the Perfect Square Trinomial 9x^2+6x+d

$9 x^{2} + 6 x + d$

Step 1

Move $6 x$ .

$9 x^{2} + d + 6 x$

Step 2

Factor out the $9$

$9 (x^{2} + \frac{2 x}{3} + \frac{d}{9})$

Step 3

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.

${(\frac{2}{3} \cdot \frac{1}{2})}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(\frac{2}{3} \cdot \frac{1}{2})}^{2}$

Rewrite the expression.

${(\frac{1}{3})}^{2}$

Simplify the expression.

Apply the product rule to $\frac{1}{3}$ .

$\frac{1^{2}}{3^{2}}$

One to any power is one.

$\frac{1}{3^{2}}$

Raise $3$ to the power of $2$ .

$\frac{1}{9}$

Step 4

Add $\frac{1}{9}$ to get the perfect square trinomial.

$9 (x^{2} + \frac{2 x}{3} + \frac{d}{9} + \frac{1}{9})$

Step 5

Simplify.

Apply the distributive property.

$9 x^{2} + 9 \frac{2 x}{3} + 9 \frac{d}{9} + 9 (\frac{1}{9})$

Simplify.

Cancel the common factor of $3$ .

Factor $3$ out of $9$ .

$9 x^{2} + 3 (3) \frac{2 x}{3} + 9 \frac{d}{9} + 9 (\frac{1}{9})$

Cancel the common factor.

$9 x^{2} + 3 \cdot 3 \frac{2 x}{3} + 9 \frac{d}{9} + 9 (\frac{1}{9})$

Rewrite the expression.

$9 x^{2} + 3 (2 x) + 9 \frac{d}{9} + 9 (\frac{1}{9})$

Multiply $2$ by $3$ .

$9 x^{2} + 6 x + 9 \frac{d}{9} + 9 (\frac{1}{9})$

Cancel the common factor of $9$ .

Cancel the common factor.

$9 x^{2} + 6 x + 9 \frac{d}{9} + 9 (\frac{1}{9})$

Rewrite the expression.

$9 x^{2} + 6 x + d + 9 (\frac{1}{9})$

Cancel the common factor of $9$ .

Cancel the common factor.

$9 x^{2} + 6 x + d + 9 (\frac{1}{9})$

Rewrite the expression.

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

Find the Perfect Square Trinomial 9x^2+6x+d

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