Solving x^4+4x^3+3x^2-20x-20=0 using substitution

[Physicsissuef]

Homework Statement

Find x.

$$x^4+4x^3+3x^2-20x-20=0$$

Homework Equations

The Attempt at a Solution

$$x^4+4x^3+3x^2-20x-20=0$$/:$$x^2$$

$$x^2+4x+3-20\frac{1}{x}-20\frac{1}{x^2}=0$$

$$(x^2-20\frac{1}{x^2})+4(x-5\frac{1}{x})+3=0$$

$$x-5\frac{1}{x}=y$$

$$x^2-10+25\frac{1}{x^2}=y^2$$

$$x^2+25\frac{1}{x^2}=y^2+10$$

? I can't substitute for $$x^2-20\frac{1}{x^2}$$

 

[colby2152]

Perhaps thehttp://www.sparknotes.com/math/algebra2/polynomials/section4.rhtml" may help you!

 

[Pere Callahan]

Sometimes it's a good idea to simply guess a solution.

Good candidates are integers and the rationals allowed by the theorem mentioned by colby.

Once you have found one solution (no matter how) you can factor out this root and are left with a polynomial of degree one less than the original one.

Can you guess any solution to your equation?

What about x=0 ? x=1 ? x=2 ? x= -1 ? x= -2 ?

 

[jimvoit]

Is it a candidate for factoring into three factors? A quadradic and two functions of x ?

 

[Dick]

Is it a candidate for factoring into three factors? A quadradic and two functions of x ?

Yes.

 

[Physicsissuef]

Yes, this problem can be solved with Perhaps the Rational Zeros Theorem. thanks for all guys.

 

[Redbelly98]

The terms "-20x - 20" jumped out at me. What value of x will make those two terms cancel? Will that same value of x make the other terms cancel out?

That's one factor down, three to go.

 

[HallsofIvy]

The next root is just as easy and then you are left with a quadratic!