How do you solve $3x^3+4x^2=75x+100$ ?

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Answer 1 · 2017-08-17T12:40:46

$x = -5, x = 5, x = -4/3$

$3x^3+ 4x^2 = 75x + 100$

$3x^3 + 4x^2 - 75x - 100 = 0$

$x^2(3x + 4) - 25 (3x + 4) = 0$

$(x^2 - 25)(3x + 4) = 0$

$(x+5)(x-5)(3x + 4) = 0$

$x + 5 = 0, x - 5 = 0, 3x + 4 = 0$

$x = -5, x = 5, x = -4/3$

How do you solve 3x^3+4x^2=75x+1003x^3+4x^2=75x+100?