# X^4 equation

## Homework Statement

Find x.

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

## 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 $$x^2-20\frac{1}{x^2}$$

Related Precalculus Mathematics Homework Help News on Phys.org
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 ?

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

Dick
Homework Helper
Is it a candidate for factoring into three factors? A quadradic and two functions of x ?
Yes.

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

Redbelly98
Staff Emeritus
Homework Helper
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