Find the Perfect Square Trinomial x^2-4/3x

$x^{2} - \frac{4}{3} x$

Step 1

Simplify each term.

Combine $x$ and $\frac{4}{3}$ .

$x^{2} - \frac{x \cdot 4}{3}$

Move $4$ to the left of $x$ .

$x^{2} - \frac{4 x}{3}$

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.

${(- \frac{4}{3} \cdot \frac{1}{2})}^{2}$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{4}{3}$ into the numerator.

${(\frac{- 4}{3} \cdot \frac{1}{2})}^{2}$

Factor $2$ out of $- 4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

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

Apply the product rule to $- \frac{2}{3}$ .

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

Apply the product rule to $\frac{2}{3}$ .

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

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

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

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

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

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

$\frac{4}{3^{2}}$

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

$\frac{4}{9}$

Step 3

Add $\frac{4}{9}$ to get the perfect square trinomial.

$x^{2} - \frac{4}{3} x + \frac{4}{9}$