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How to find roots of polynomials

Integer roots
If the coefficients of a polynomial are integers, it is natural to look for roots which are also integers. Any such root must divide the constant term. We can often ``guess" one or more roots by trying all possibilities.


If there is an integer root, it must divide -2. This leaves only four possibilities: 1, -1, 2, and -2. By plugging in, we find that

1^4-2\cdot 1^2-3\cdot 1-2=-5,\end{displaymath}


2^4-2\cdot 2^2-3\cdot 2-2=0,\end{displaymath}


Therefore -1 and 2 are roots, but 1 and -2 are not.
Reducing the degree
If a root has been found, you can reduce the degree of the polynomial to get a simpler problem for the remaining roots. For example, you know from above that


has the roots -1 and 2. This tells you that the polynomial must contain the factors r+1 and r-2. You can use the long division algorithm to find




Therefore, the remaining roots must solve


The quadratic formula gives

r=-{1\over 2}\pm i{\sqrt{3}\over 2}.\end{displaymath}

In summary, we have found that the roots of


are -1, 2, $-1/2+i\sqrt{3}/2$ and $-1/2-i\sqrt{3}/2$.
Multiple roots
It can happen that a polynomial contains the same linear factor more than once. For instance,


Since the factor r-1 appears twice, we call 1 a double root of the polynomial, while -1 is a simple root. Note that

{d\over dr}(r^3-r^2-r+1)=3r^2-2r-1\end{displaymath}

also has a root at r=1. This happens in general: If a polynomial P(r) as a k-fold root at r=c, then



P^{(k)}(c)\neq 0.\end{displaymath}

If roots are counted by multiplicity (i.e. a double root counts twice, a triple root three times etc.), then a polynomial of nth degree has n roots.
The nth root of a complex number
We want to find the roots of the equation


where w is a given number. w may be complex, but the following procedure is important even if w is real. The solution of the equation requires writing w in polar form


That is, if x and y are the real an imaginary parts of w, we want to find $\rho$ and $\theta$ is such a way that $x=\rho\cos\theta$ and $y=\rho\sin\theta$. In other words, $\rho$ and $\theta$ are polar coordinates of the point (x,y) in the Cartesian plane.
1. w=1. In this case x=1 and y=0. We can choose $\rho=1$ and $\theta=0$.
2. w=-1. Now x=-1 and y=0. We have $\rho=1$ and $\theta=\pi$.
3. w=1+i. We have $\rho=\sqrt{2}$ and $\theta=\pi/4$.
The point of using polar form is that it is very easy to take the nth root. We find


Since $\theta$ is an angle, it is not unique; it is determined only up to a multiple of $2\pi$. That is, in the polar form of w, we could have written


instead of $w=\rho\exp(i\theta)$. Although, $\exp(i\theta)=\exp(i(\theta+2\pi))$we have

\exp(i\theta/n)\neq \exp(i(\theta+2\pi)/n)\end{displaymath}

if n>1. To find all nth roots of w, you must consider the representations


where k takes the integer values , 1, ..., n-1 (if you go further, then k=n will give you the same as k=0). The nth roots of w are then




We put -1 into polar form


For the third roots, we find

r_1=\exp(i\pi/3)=\cos(\pi/3)+i\sin(\pi/3)={1\over 2}+i{\sqrt{3}\over 


r_3=\exp(5i\pi/3)=\cos(5\pi/3)+i\sin(5\pi/3)={1\over 2}-i{\sqrt{3}\over 

We could have found the same result from the factorization


The quadratic formula gives

{1\over 2}\pm i{\sqrt{3}\over 2}\end{displaymath}

for the roots of the second factor.
Beware of the difference between




The first equation has a triple root at -1, the second has three different roots: -1 and $1/2\pm i\sqrt{3}/2$. In general, the equation rn=w always has n DIFFERENT ROOTS. Be prepared for a whipping with the cat of nine tails if you should ever say that the equation rn=w has an n-fold root.
next up previous
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Michael Renardy