# How do you use the perfect square trinomial formula to factor 16y^2-8y+1?

Aug 4, 2018

$16 {y}^{2} - 8 y + 1 = \left(4 y - 1\right) \left(4 y - 1\right)$

#### Explanation:

Here,
$f \left(y\right) = 16 {y}^{2} - 8 y + 1$
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To complete perfect square let $M$ be the third term,
such that $16 {y}^{2} - 8 y + M$ is a perfect square.

So,

color(blue)((i)1^(st) term=16y^2
color(blue)((ii)2^(nd)term=-8y
color(blue)((iii)3^(rd)term=M

Formula :
color(blue)(3^(rd)term=(2^(nd)term)^2/(4xx 1^(st) term)

$M = {\left(- 8 y\right)}^{2} / \left(4 \times 16 {y}^{2}\right) = \frac{64 {y}^{2}}{64 {y}^{2}} = 1$

$i . e . 16 {y}^{2} - 8 y + 1$ is a perfect square.
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So ,

$f \left(y\right) = 16 {y}^{2} - 8 y + 1$

$\therefore f \left(y\right) = {\left(4 y\right)}^{2} - 2 \left(4 y\right) \left(1\right) + {\left(1\right)}^{2}$

$\therefore f \left(y\right) = {\left(4 y - 1\right)}^{2}$