Find the Perfect Square Trinomial k^2-9k

$k^{2} - 9 k$

Step 1

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

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{9}{2}$ .

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

Apply the product rule to $\frac{9}{2}$ .

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

Raise $- 1$ to the power of $2$ .

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

Multiply $\frac{9^{2}}{2^{2}}$ by $1$ .

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

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

$\frac{81}{2^{2}}$

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

$\frac{81}{4}$

Step 2

Add $\frac{81}{4}$ to get the perfect square trinomial.

$k^{2} - 9 k + \frac{81}{4}$