# X^4 equation

1. Apr 13, 2008

### Physicsissuef

1. The problem statement, all variables and given/known data
Find x.

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

2. Relevant equations

3. 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 $$x^2-20\frac{1}{x^2}$$

2. Apr 13, 2008

### colby2152

Last edited by a moderator: Apr 23, 2017
3. Apr 14, 2008

### 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 ?

4. Apr 14, 2008

### jimvoit

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

5. Apr 14, 2008

### Dick

Yes.

6. Apr 15, 2008

### Physicsissuef

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

7. Apr 17, 2008

### Redbelly98

Staff Emeritus
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.

8. Apr 18, 2008

### HallsofIvy

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