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

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Answer 1 · 2018-07-20T04:17:24

$36 x^{2} + 35 x + 8 = (72 x + 35 - \sqrt{73}) (72 + 35 + \sqrt{73})$ $36x^2+35x+8=(72x+35-sqrt73)(72+35+sqrt73)$

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

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

$x = \frac{- 35 \pm \sqrt{35^{2} - 4 \times 36 \times 8}}{2 \times 36}$ $x=(-35+-sqrt(35^2-4times36times8))/(2times36)$

$x = \frac{- 35 \pm \sqrt{73}}{72}$ $x=(-35+-sqrt73)/72$

$x = \frac{- 35 + \sqrt{73}}{72}$ $x=(-35+sqrt73)/72$ or $x = \frac{- 35 - \sqrt{73}}{72}$ $x=(-35-sqrt73)/72$

$36 x^{2} + 35 x + 8$ $36x^2+35x+8$

$= (x - (\frac{- 35 + \sqrt{73}}{72})) (x - \frac{- 35 - \sqrt{73}}{72}))$ $=(x-((-35+sqrt73)/72))(x-(-35-sqrt73)/72))$

$= (x + \frac{35}{72} - \frac{\sqrt{73}}{72}) (x + \frac{35}{72} + \frac{\sqrt{73}}{72})$ $=(x+35/72-sqrt73/72)(x+35/72+sqrt73/72)$

$= (72 x + 35 - \sqrt{73}) (72 + 35 + \sqrt{73})$ $=(72x+35-sqrt73)(72+35+sqrt73)$

How do you factor 36x2+35x+836x^2 + 35x + 8?