How do you factor $16y^2 - 9$ ?

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Answer 1 · 2018-03-03T16:02:27

See a solution process below:

This is a special form of the quadratic:

$color(red)(a)^2 - color(blue)(b)^2 = (color(red)(a) + color(blue)(b))(color(red)(a) - color(blue)(b))$

Let: $a^2 = 16y^2$ then:

$sqrt(a^2) = sqrt(16y^2)$

$a = 4y$

Let: $b^2 = 9$ then:

$sqrt(b^2) = sqrt(9)$

$b = 3$

Substituting into the rule gives:

$16y^2 - 9 => color(red)((4y))^2 - color(blue)(3)^2 => (color(red)(4y) + color(blue)(3))(color(red)(4y) - color(blue)(3))$

Answer 2 · 2018-03-03T16:06:38

$(4y+3)(4y-3)$

$color(red)(A^2-B^2=(A+B)(A-B))$
$16y^2-9=(4y)^2-(3)^2$
Applying above formula,
$16y^2-9=(4y+3)(4y-3)$

How do you factor 16y^2 - 916y^2 - 9?