How do you factor $v^{4} + 16 v^{2} d^{2} + 64 d^{4}$ $v^ { 4} + 16v ^ { 2} d ^ { 2} + 64d ^ { 4}$ ?

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How do you factor $v^{4} + 16 v^{2} d^{2} + 64 d^{4}$ $v^ { 4} + 16v ^ { 2} d ^ { 2} + 64d ^ { 4}$ ?

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Answer 1 · 2017-12-05T20:43:57

${(v^{2} + 8 d^{2})}^{2}$ $(v^2+8d^2)^2$

Explanation:

$v^{4} + 16 v^{2} d^{2} + 64 d^{4}$ $v^4+16v^2d^2+64d^4$

= ${(v^{2} + 8 d^{2})}^{2}$ $(v^2+8d^2)^2$

Answer 2 · 2017-12-05T20:44:55

${(v^{2} + 8 d^{2})}^{2}$ $(v^2+8d^2)^2$

Explanation:

We can recognize that the expression is in the form of the expansion of ${(a + b)}^{2}$ $(a+b)^2$ :
${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $(a+b)^2=a^2+2ab+b^2$

We can therefor tell that $a^{2} = v^{4}$ $a^2=v^4$ and $b^{2} = 64 d^{4}$ $b^2=64d^4$ . Taking the square root of both of these equations, we get:
$a = v^{2}$ $a=v^2$ , $b = 8 d^{2}$ $b=8d^2$ (Note that we're only interested in the positive square root, since the middle coefficient is positive).

This is all we need to know to be able to factor like so:
$v^{4} + 16 v^{2} d^{2} + 64 d^{4} = {(v^{2} + 8 d^{2})}^{2}$ $v^4+16v^2d^2+64d^4=(v^2+8d^2)^2$

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How do you factor $v^{4} + 16 v^{2} d^{2} + 64 d^{4}$ $v^ { 4} + 16v ^ { 2} d ^ { 2} + 64d ^ { 4}$ ?

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Explanation:

Explanation:

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How do you factor $v^{4} + 16 v^{2} d^{2} + 64 d^{4}$ $v^ { 4} + 16v ^ { 2} d ^ { 2} + 64d ^ { 4}$ ?