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

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

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Answer 1 · 2018-03-03T16:02:27

See a solution process below:

Explanation:

This is a special form of the quadratic:

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

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

$\sqrt{a^{2}} = \sqrt{16 y^{2}}$ $sqrt(a^2) = sqrt(16y^2)$

$a = 4 y$ $a = 4y$

Let: $b^{2} = 9$ $b^2 = 9$ then:

$\sqrt{b^{2}} = \sqrt{9}$ $sqrt(b^2) = sqrt(9)$

$b = 3$ $b = 3$

Substituting into the rule gives:

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

Answer 2 · 2018-03-03T16:06:38

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

Explanation:

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

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

2 Answers

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