Find the Perfect Square Trinomial t^2-14/3t+n

$t^{2} - \frac{14}{3} t + n$

Step 1

Simplify each term.

Combine $t$ and $\frac{14}{3}$ .

$t^{2} - \frac{t \cdot 14}{3} + n$

Move $14$ to the left of $t$ .

$t^{2} - \frac{14 t}{3} + n$

Step 2

Move $- \frac{14 t}{3}$ .

$t^{2} + n - \frac{14 t}{3}$

Step 3

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

Multiply the numerator by the reciprocal of the denominator.

${(- \frac{14}{3} \cdot \frac{1}{2})}^{2}$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{14}{3}$ into the numerator.

${(\frac{- 14}{3} \cdot \frac{1}{2})}^{2}$

Factor $2$ out of $- 14$ .

${(\frac{2 (- 7)}{3} \cdot \frac{1}{2})}^{2}$

Cancel the common factor.

${(\frac{2 \cdot - 7}{3} \cdot \frac{1}{2})}^{2}$

Rewrite the expression.

${(\frac{- 7}{3})}^{2}$

Move the negative in front of the fraction.

${(- \frac{7}{3})}^{2}$

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

Apply the product rule to $- \frac{7}{3}$ .

${(- 1)}^{2} {(\frac{7}{3})}^{2}$

Apply the product rule to $\frac{7}{3}$ .

${(- 1)}^{2} \frac{7^{2}}{3^{2}}$

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

$1 \frac{7^{2}}{3^{2}}$

Multiply $\frac{7^{2}}{3^{2}}$ by $1$ .

$\frac{7^{2}}{3^{2}}$

Raise $7$ to the power of $2$ .

$\frac{49}{3^{2}}$

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

$\frac{49}{9}$

Step 4

Simplify each term.

Combine $t$ and $\frac{14}{3}$ .

$t^{2} - \frac{t \cdot 14}{3} + n + \frac{49}{9}$

Move $14$ to the left of $t$ .

$t^{2} - \frac{14 t}{3} + n + \frac{49}{9}$