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# Solutions to polynomials of high degrees

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.

Note by Luke Zhang
1 year, 10 months ago

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You 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? Staff · 1 year, 10 months ago

Yea. This can only work for integers. · 1 year, 10 months ago

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. Staff · 1 year, 10 months ago

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

Try for $x^{3} - 7x + 7$ · 1 year, 10 months ago

You can use newton sums to bash.............................................. · 1 year, 10 months ago

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 . · 1 year, 10 months ago

So what I would do is just factor -7 into -1 and 7 Sub it into the LHS such that x=-1 and (x^2)-7 = 7. So to see if this works. Make sure that you can solve for both sides. Such that when I sub x as -1, (x^2)-7 must equal to 7. However this would not work for your equation as x must be integer. · 1 year, 10 months ago

Ok ,thanks :) · 1 year, 10 months ago