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)#