Is there any way I can find the roots to these easily or is it just guessing and checking?

$\displaystyle r^3 + r^2 - 2 = 0$

I also what to find the roots to:

$\displaystyle r^3 + 3r^2 + 3r + 1 = 0$

Any help would be appreciated.

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- Feb 24th 2011, 04:55 PMDarKFinding roots
Is there any way I can find the roots to these easily or is it just guessing and checking?

$\displaystyle r^3 + r^2 - 2 = 0$

I also what to find the roots to:

$\displaystyle r^3 + 3r^2 + 3r + 1 = 0$

Any help would be appreciated. - Feb 24th 2011, 04:58 PMmr fantastic
- Feb 25th 2011, 04:03 AMHallsofIvy
If it was not quite "obvious" enough for you, use the "rational root" theorem: If the rational number $\displaystyle \frac{m}{n}$ is a root of a polynomial equation, with integer coefficients, then the integer "n" must evenly divide the leading coefficient and the integer "m" must evenly divide the constant term.

For this equation, $\displaystyle r^3+ r^2- 2= 0$ the leading coefficient is 1 and the constant term is -2. The only integers that divide 1 are 1 and -1 and the only integers that divide -2 are 1, -1, 2, and -2. That tells you that the only possible rational roots are $\displaystyle \frac{1}{1}= 1$, $\displaystyle \frac{-1}{1}= -1$, $\displaystyle \frac{2}{1}= 2$ and $\displaystyle \frac{-2}{1}= -2$ (there are other combinations but they give the same results). Putting those into the equation shows that r= 1 is the only rational root. Divide the polynomial $\displaystyle r^3+ r^2- 2$ by x- 1, as mr fantastic suggests, to get a quadratic equation the other two roots must satisfy. Since none of -1, 2, or -2 satisfy the orginal equation, this quadratic equation will have no rational roots. Solve by either completing the square or using the quadratic formula.

You can try the same sort of thing with $\displaystyle r^3+ 3r^2+ 3r+ 1= 0$. Now both leading coefficient and constant term are 1 so the only possible rational roots are 1 and -1. r= 1 obviously does not give 0 but r= -1 gives -1+ 3- 3+ 1= 0 so -1 is a root. Again, divide the polynomial by r-(-1)= r+ 1 to get a quadratic polynomial. It will turn out to have two rational roots so, by what I have said, you should be able guess what they are! - Mar 14th 2011, 07:07 PMDarK
Thank you so much, I tried it and so far it is helping me ALOT!

So in general, the rational root theorem will work for higher degree polynomials as well? ie. A polynomial that is 4th degree, 8th, etc.. - Mar 14th 2011, 07:14 PMProve It
- Mar 15th 2011, 10:43 AMtopsquark
As long as the coefficients of the y^n terms are integers.

$\displaystyle \displaystyle \frac{1}{2}y^4 - 3y^2 - y + 3 = 0$

can be transformed into

$\displaystyle \displaystyle y^4 - 6y^2 - 2y + 6 = 0$

by multiplying through by 2. So the rational root theorem can be applied, though there may not be any rational roots to the equation. But the equation

$\displaystyle \displaystyle y^3 + \sqrt{2} y + 1 = 0$

cannot be solved by the rational root theorem. (The good news is you probabloy won't run into any of these.)

-Dan - Mar 16th 2011, 03:10 AMHallsofIvy
We should also note another point- a polynomial with integer coefficients may

**not**have any rational roots at all. In that case it would probably be very difficult to solve the equation.