Find the Perfect Square Trinomial 4x^2+4x+1

$4 x^{2} + 4 x + 1$

Factor out the $4$

$4 (x^{2} + x + \frac{1}{4})$

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

Apply the product rule to $\frac{1}{2}$ .

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

One to any power is one.

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

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

$\frac{1}{4}$

Add $\frac{1}{4}$ to get the perfect square trinomial.

$4 (x^{2} + x + \frac{1}{4} + \frac{1}{4})$

Simplify.

Combine the numerators over the common denominator.

$4 (x^{2} + x + \frac{1 + 1}{4})$

Add $1$ and $1$ .

$4 (x^{2} + x + \frac{2}{4})$

Cancel the common factor of $2$ and $4$ .

Factor $2$ out of $2$ .

$4 (x^{2} + x + \frac{2 (1)}{4})$

Cancel the common factors.

Factor $2$ out of $4$ .

$4 (x^{2} + x + \frac{2 \cdot 1}{2 \cdot 2})$

Cancel the common factor.

$4 (x^{2} + x + \frac{2 \cdot 1}{2 \cdot 2})$

Rewrite the expression.

$4 (x^{2} + x + \frac{1}{2})$

Apply the distributive property.

$4 x^{2} + 4 x + 4 (\frac{1}{2})$

Cancel the common factor of $2$ .

Factor $2$ out of $4$ .

$4 x^{2} + 4 x + 2 (2) \frac{1}{2}$

Cancel the common factor.

$4 x^{2} + 4 x + 2 \cdot 2 \frac{1}{2}$

Rewrite the expression.

$4 x^{2} + 4 x + 2$