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.

ax5+bx4+cx3+dx2+ex+f=0{ ax }^{ 5 }+{ bx }^{ 4 }+{ cx }^{ 3 }+{ dx }^{ 2 }+ex+f=0

Where a,b,c,d,e,fa,b,c,d,e,f are some integer.

To solve for xx, bring ff to the other side of the equation and factor out x.

This will yield,

x(ax4+bx3+cx2+dx+e)=fx({ ax }^{ 4 }+{ bx }^{ 3 }+{ cx }^{ 2 }+{ dx }+e)=-f

Finding out the factors of f-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
4 years, 3 months ago

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@Luke Zhang

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

Try for x37x+7x^{3} - 7x + 7

Azhaghu Roopesh M - 4 years, 3 months ago

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

Julian Poon - 4 years, 3 months ago

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

x37x+7=0x(x27)=71x^{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 .

Azhaghu Roopesh M - 4 years, 3 months ago

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@Azhaghu Roopesh M 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.

Luke Zhang - 4 years, 3 months ago

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@Luke Zhang Ok ,thanks :)

Azhaghu Roopesh M - 4 years, 3 months ago

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@Azhaghu Roopesh M No problem.

Luke Zhang - 4 years, 3 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 x2+x=5 x^2 +x = 5 ? How are you going to factorize 5?

Calvin Lin Staff - 4 years, 3 months ago

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Yea. This can only work for integers.

Luke Zhang - 4 years, 3 months ago

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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.

Calvin Lin Staff - 4 years, 3 months ago

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