# Help solving a third degree polynomial

1. Oct 1, 2008

### jeffmarina

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

x^2(4x + 13) = 9

2. Relevant equations

3. The attempt at a solution

Answers are that x = -3, -1 and 3/4. But how can one compute this other than trial and error?

2. Oct 1, 2008

### gabbagabbahey

First step is to try expanding it into the form $Ax^3 +Bx^2+Cx+D=0$. What do you get when you do this?

3. Oct 1, 2008

### jeffmarina

4x^3 + 13x^2 +0x -9 = 0

4. Oct 1, 2008

### gabbagabbahey

Okay good, you can get rid of the 0x term though :)

The next step is to recognize that since this is a third degree polynomial, there will be three roots; let's call them $x_1$, $x_2$, and $x_3$. This means that $4(x-x_1)(x-x_2)(x-x_3)=0=4x^3+13x^2-9$. Expand the expression on the left. What do you get?

5. Oct 1, 2008

### HallsofIvy

Staff Emeritus
If you did not already know the solutions, you could use the "rational root theorem": if x= m/n is a rational number satisfying a polynomial equation then n must evenly divide the leading coefficient and m must evenly divide the constant term.

Here, those are 4 and -9. The integers that evenly divide 4 are $\pm 1$, $\pm 2$ and $\pm 4$. The integers that evenly divide 9 are $\pm 1$, $\pm 3$, and $\pm 9$. That means that the only possible rational roots are $\pm 1$, $\pm 1/2$, $\pm 1/4$, $\pm 3$, $\pm 3/2$, $\pm 3/4$, $\pm 9$, $\pm 9/2$, and $\pm 9/4$. Check each of those to see if one of them satisfies the equation. (And notice that the solutions you give are among those.)

See mathworld.wolfram.com/CubicFormula.html

Of course, there exist cubic equations that have no rational roots. In such a case, you would have to use the (extremely complicated) cubic formula.

6. Oct 1, 2008

### Defennder

Hi, gabba just want to make sure I understand your method here. Is it possible that this may not work for some polynomials since we may end up with a cubic equation of a variable say x_1 ? Which of course leads us back to square one.

7. Oct 1, 2008

### gabbagabbahey

Yes, you bring up a good point that I had neglected to mention. If the polynomial in question happens to be a perfect cube, then this method won't get you anywhere. However if it is a perfect cube, when you put the polynomial into the form $x^3+C_1x^2+C_2x+C_3$, the constant term $C_3$ will be the cube of the triple root as we can see from the fact that $(x-a)^3=x^3-3ax^2+3a^2x-a^3$.

8. Oct 2, 2008

### Defennder

But is this guaranteed to work if the polynomial isn't a perfect cube ? ie. One would only have to work with quadratics at most ?