Cardano of Milan and Ferrari(His Student) solved the cubic and quartic equation.

View Cardano's Method

I find it way easier(other than using the cubic formula) to just use a simple algorithm to solve for the polynomial. Note that this approach only works when the cubic equation has an integer root. It is a result of (but not equivalent to) the rational root theorem.

For example, given a quintic equation.

\({ ax }^{ 5 }+{ bx }^{ 4 }+{ cx }^{ 3 }+{ dx }^{ 2 }+ex+f=0\)

Where \(a,b,c,d,e,f\) are some integer.

To solve for \(x\), bring \(f\) to the other side of the equation and factor out x.

This will yield,

\(x({ ax }^{ 4 }+{ bx }^{ 3 }+{ cx }^{ 2 }+{ dx }+e)=-f\)

Finding out the factors of \(-f\)(To 2 factors)

And substituting in the values into the equation.

Make sure both sides works for the equation. If so then you will find the solution.

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## Comments

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TopNewestYou are assuming that there must be an integer (or even rational) solution. How do you propose to use your method to solve \( x^2 +x = 5 \)? How are you going to factorize 5?

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Yea. This can only work for integers.

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Right. As such, you do not have a general solution like Cardano. What you have is "a method that works under a very specific set of instances".

In fact, what you have is the Rational Root Theorem.

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@Luke Zhang

Hi , can you provide an example to prove your theory ?

Try for \[x^{3} - 7x + 7\]

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You can use newton sums to bash..............................................

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Agreed , but I want to know what he (Luke) wanted to convey by writing this note ?

\[x^{3} - 7x + 7 =0 \\ x(x^{2} -7 ) = -7\cdot 1 \]

Now what , does he want to input -7 and 1 into the LHS , what good will it do ? This is what I am asking .

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