Find the Perfect Square Trinomial x^2-19x

$x^{2} - 19 x$

To find the $c$ value $c = {(\frac{b}{2})}^{2}$ , divide the coefficient of $x$ 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 $- \frac{19}{2}$ .

${(- 1)}^{2} {(\frac{19}{2})}^{2}$

Apply the product rule to $\frac{19}{2}$ .

${(- 1)}^{2} \frac{19^{2}}{2^{2}}$

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

$1 \frac{19^{2}}{2^{2}}$

Multiply $\frac{19^{2}}{2^{2}}$ by $1$ .

$\frac{19^{2}}{2^{2}}$

Raise $19$ to the power of $2$ .

$\frac{361}{2^{2}}$

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

$\frac{361}{4}$

Add $\frac{361}{4}$ to get the perfect square trinomial.

$x^{2} - 19 x + \frac{361}{4}$