# math

solve the equation x^3-5x^2-4x+20=0

I need a hint on how to solve this

Look for the first root by graphical means or trial-and-error. Note that when x=2,
x^3-5x^2-4x+20 = 8-20 -8 +20 = 0.
Therefore 2 is one root and x-2 is a factor.
Next, divide x-2 into x^3-5x^2-4x+20 using long division, to get a quadratic factor. The result is
(x-2)(x^2-3x-10)
The quadratic factor is just (x-5)(x+2), so the equation becomes
(x-2)(x+2)(x-5) = 0
This tells you that the other roots are x = -2 and +5.

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