Determine if the Expression is a Perfect Square 16x^2-8x+1

$16 x^{2} - 8 x + 1$

A trinomial can be a perfect square if it satisfies the following:

The first term is a perfect square.

The third term is a perfect square.

The middle term is either $2$ or $- 2$ times the product of the square root of the first term and the square root of the third term.

${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$

Find $a$ , which is the square root of the first term $16 x^{2}$ . The square root of the first term is $\sqrt{16 x^{2}} = 4 x$ , so the first term is a perfect square.

Rewrite $16 x^{2}$ as ${(4 x)}^{2}$ .

$\sqrt{{(4 x)}^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$4 x$

Any root of $1$ is $1$ .

$1$

The first term $16 x^{2}$ is a perfect square. The third term $1$ is a perfect square. The middle term $- 8 x$ is $- 2$ times the product of the square root of the first term $4 x$ and the square root of the third term $1$ .

The polynomial is a perfect square. ${(4 x - 1)}^{2}$

Determine if the Expression is a Perfect Square 16x^2-8x+1

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