How do you factor $v^{2} + 21 v + 30$ $v^2 +21v +30$ ?

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How do you factor $v^{2} + 21 v + 30$ $v^2 +21v +30$ ?

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Answer 1 · 2017-02-20T23:31:14

$v^{2} + 21 v + 30 = (v + \frac{21}{2} - \frac{\sqrt{321}}{2}) (v + \frac{21}{2} + \frac{\sqrt{321}}{2})$ $v^2+21v+30 = (v+21/2-sqrt(321)/2)(v+21/2+sqrt(321)/2)$

Explanation:

Complete the square then use the difference of squares identity:

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

with $a = 2 v + 21$ $a=2v+21$ and $b = \sqrt{321}$ $b=sqrt(321)$ as follows:

$v^{2} + 21 v + 30 = \frac{1}{4} (4 v^{2} + 84 v + 120)$ $v^2+21v+30 = 1/4(4v^2+84v+120)$

$v^{2} + 21 v + 30 = \frac{1}{4} ({(2 v)}^{2} + 2 (2 v) (21) + {(21)}^{2} - 321)$ $color(white)(v^2+21v+30) = 1/4((2v)^2+2(2v)(21)+(21)^2-321)$

$v^{2} + 21 v + 30 = \frac{1}{4} ({(2 v + 21)}^{2} - {(\sqrt{321})}^{2})$ $color(white)(v^2+21v+30) = 1/4((2v+21)^2-(sqrt(321))^2)$

$v^{2} + 21 v + 30 = \frac{1}{4} ((2 v + 21) - \sqrt{321}) ((2 v + 21) + \sqrt{321})$ $color(white)(v^2+21v+30) = 1/4((2v+21)-sqrt(321))((2v+21)+sqrt(321))$

$v^{2} + 21 v + 30 = \frac{1}{4} (2 v + 21 - \sqrt{321}) (2 v + 21 + \sqrt{321})$ $color(white)(v^2+21v+30) = 1/4(2v+21-sqrt(321))(2v+21+sqrt(321))$

$v^{2} + 21 v + 30 = (v + \frac{21}{2} - \frac{\sqrt{321}}{2}) (v + \frac{21}{2} + \frac{\sqrt{321}}{2})$ $color(white)(v^2+21v+30) = (v+21/2-sqrt(321)/2)(v+21/2+sqrt(321)/2)$

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How do you factor $v^{2} + 21 v + 30$ $v^2 +21v +30$ ?

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