Solutions to polynomials of high degrees

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

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.

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
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Comments

@Luke Zhang

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

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

A Former Brilliant Member - 6 years, 3 months ago

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

Julian Poon - 6 years, 3 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 .

A Former Brilliant Member - 6 years, 3 months ago

@A Former Brilliant Member – 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 - 6 years, 3 months ago

@Luke Zhang – Ok ,thanks :)

A Former Brilliant Member - 6 years, 3 months ago

@A Former Brilliant Member – No problem.

Luke Zhang - 6 years, 3 months ago

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?

Calvin Lin Staff - 6 years, 3 months ago

Yea. This can only work for integers.

Luke Zhang - 6 years, 3 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.

Calvin Lin Staff - 6 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$