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solve x^3-4x^2+5x+4=0
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Since there are no rational factors. I guess I should use the Newton's Raphson Method to find the roots. Is that correct?
You could use that, OR use this equation to find it: \[ay^3+(c-\frac{b^2}{3a})y+(d+\frac{2b^3}{27a^3}-\frac{bc}{3a})=0\]
That is the \(Cubic\) formula, similar to the quadratic formula, but for \(x^3\)
I didn't know there is a cubic formula. Thanks for the info
No problem @junyang96 :)
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Actually, to my knowledge, that is actually the wrong equation @junyang96 ;). This is the right equation:
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