How do you factor $16 r^{2} - 25 a^{2}$ $16r^2 - 25a^2$ ?

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How do you factor $16 r^{2} - 25 a^{2}$ $16r^2 - 25a^2$ ?

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Answer 1 · 2018-04-25T19:19:12

$16 r^{2} - 25 a^{2} = (4 r + 5 a) (4 r - 5 a)$ $16r^2-25a^2=color(blue)((4r+5a)(4r-5a))$

Explanation:

Remember: $(x^{2} - y^{2})$ $(x^2-y^2)$ can be factored as $(x + y) (x - y)$ $(x+y)(x-y)$

Using $16 r^{2}$ $16r^2$ as $x^{2}$ $x^2$ (that is using $4 r$ $4r$ as $x$ $x$ ), and
using $25 a^{2}$ $25a^2$ as $y^{2}$ $y^2$ (that is using $5 a$ $5a$ as $y$ $y$ )

we get
$XXX (4 r + 5 a) (4 r - 5 a)$ $color(white)("XXX")(4r+5a)(4r-5a)$

Answer 2 · 2018-04-25T19:22:19

$(4 r - 5 a) (4 r + 5 a)$ $(4r-5a)(4r+5a)$

Explanation:

$16 r^{2} - 25 a^{2} is a difference of squares$ $16r^2-25a^2" is a "color(blue)"difference of squares"$

$which factors in general as$ $"which factors in general as"$

$∙ x a^{2} - b^{2} = (a - b) (a + b)$ $•color(white)(x)a^2-b^2=(a-b)(a+b)$

$here 16 r^{2} = {(4 r)}^{2} and 25 a^{2} = {(5 a)}^{2}$ $"here "16r^2=(4r)^2" and "25a^2=(5a)^2$

$\Rightarrow a = 4 r and b = 5 a$ $rArra=4r" and "b=5a$

$\Rightarrow 16 r^{2} - 25 a^{2} = (4 r - 5 a) (4 r + 5 a)$ $rArr16r^2-25a^2=(4r-5a)(4r+5a)$

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How do you factor $16 r^{2} - 25 a^{2}$ $16r^2 - 25a^2$ ?

2 Answers

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How do you factor $16 r^{2} - 25 a^{2}$ $16r^2 - 25a^2$ ?