How do you factor $36x^2 + 35x + 8$ ?

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Answer 1 · 2018-07-20T04:17:24

$36x^2+35x+8=(72x+35-sqrt73)(72+35+sqrt73)$

$x=(-b+-sqrt(b^2-4ac))/(2a)$ where the equation is $ax^2+bx+c=0$

Therefore,
$a=36$ , $b=35$ and $c=8$

$x=(-35+-sqrt(35^2-4times36times8))/(2times36)$

$x=(-35+-sqrt73)/72$

$x=(-35+sqrt73)/72$ or $x=(-35-sqrt73)/72$

$36x^2+35x+8$

$=(x-((-35+sqrt73)/72))(x-(-35-sqrt73)/72))$

$=(x+35/72-sqrt73/72)(x+35/72+sqrt73/72)$

$=(72x+35-sqrt73)(72+35+sqrt73)$

How do you factor 36x^2 + 35x + 836x^2 + 35x + 8?