# Solve an equation

1. Sep 15, 2011

### Gavroy

hi,

i want to solve the equation
x³+x²+x-a=0
analytically
i was wondering if there are other methods than the general formula(Cardano's equation) to do this.

what i already noticed is that one can rewrite this as b=(1-x^4)/(1-x)

but i do not know if this could be helpful, do you have any further ideas?

2. Sep 15, 2011

### Staff: Mentor

What is the context of the question? Where does it come from? Is it schoolwork?

3. Sep 15, 2011

### Gavroy

no, i was curious about this, cause we solved this type of equation with newton's method and i thought maybe one could do this analytically and i know about the general formula to solve third order equation, but maybe there is another way than cardano's equations, especially as one could rewrite this to a different type of equation.

4. Sep 15, 2011

### gb7nash

Nope! The only methods you can really rely on are:

1) rational root theorem, depending if a is rational. Using this method, you can derive all of the possible rational roots and just test each one. If the rational root test fails (all numbers you test != 0) you can conclude there there are no rational roots.

2) cubic formula

3) iterative methods (newton's,bisection,etc.)

4) If you know one exact root (call it r), you can do polynomial division, dividing the cubic by (x-r) and obtaining a quadratic equation. After this, just use the quadratic formula to obtain the other two roots.

5. Sep 15, 2011

### Gavroy

the thing with the rational fractions sounds interesting.

my equation is:

0=x+x²+x³-38/27

and the rational number 2/3 is a solution.

but how do i get this by this theorem?

6. Sep 15, 2011

### gb7nash

First, we have to convert this to an integer polynomial, so we multiply both sides by 27:

0=27x³+27x²+27x-38

We now list all p/q such that p is an integer factor of -38 and q is an integer factor of 27, so we have:

$$\pm \frac{1}{1}, \pm \frac{1}{3}, \pm \frac{1}{9}, \pm \frac{1}{27}, \pm \frac{2}{1}, \pm \frac{2}{3}, \pm \frac{2}{9}, \pm \frac{2}{27}, \pm \frac{19}{1}, \pm \frac{19}{3}, \pm \frac{19}{9}, \pm \frac{19}{27}, \pm \frac{38}{1}, \pm \frac{38}{3}, \pm \frac{38}{9}, \pm \frac{38}{27}$$

Now the fun part. You plug each one into the cubic and see if any of them are roots. Out of this list is 2/3, which is a root.

7. Sep 16, 2011

### Gavroy

okay, this is really what i was looking for, thank you. do you know if there is a theorem that tells me if a polynomial has a rational number that makes it zero?

8. Sep 16, 2011

### gb7nash

Yes...the rational root theorem. If the rational root theorem fails, you can conclude that you do not have a rational root.

9. Sep 16, 2011

### Nanas

The Rational Zero theorem : it says

Suppose a polynomial
$f(x) = a_{n}x^{n} + a_{n-1}x^{n-1} + ... + a_{2}x^{2} + a_{1}x + a_{0}$
where n is natural number where $n \geq 1$ ,$a_{n} ,a_{n-1} ,...,a_{2}, a_{1}, a_{0}$ are integers.If r is a rational zero then r has the form $\pm \frac{p}{q}$ , such that $q$ is a factor of $a_{0}$ and $p$ is a factor of $a_{n}$.

you can prove the theorem by assuming $\frac{p}{q}$ is a rational zero in lowest terms of f(x),and then plug it in f(x) where $f(\frac{p}{q})=0$ , after that multiply both sides of the equation by $q^{n}$ ,finally subtract $a_{n}p^{n}$from both sides and multiply by (-1) and then make a good use of the fact that if a is a factor of one of the sides of the equation then it must be a factor of the other side. i.e that $a_{n}p^{n}$ is a multiple of p and since the other side is a multiple of q , then $a_{n}p^{n}$ is a multiple of q , but since q and p has no common factors then it follows that $a_{n}$ is a multiple of q , similarly we can show that $a_{0}$ is a multiple of p

Last edited: Sep 16, 2011