Hi,
If it is already factored like that you can see that the m^3 term causes three zero solutions.
If not, you can divide by the known roots and reduce the equation to a lower order equation and go from there.
For example, since a root is -1 divide by (m+1) first, then for -2 divide by (m+2), etc.
There might be another trick out there too.
For example:
m^5+11*m^4+46*m^3+92*m^2+88*m+32=0
gives solutions -1, -2, and -4.
Dividing that by (m+1), it goes into it evenly with no remainder. Dividing by (m+1) a second time will show a remainder which means -1 is only a solution one time.
Dividing by (m+4) will do the same thing as (m+1) so -4 is a solution only once also.
If you did (m+2) before (m+4) however, you would find that (m+2) goes into that equation more than one time without causing a remainder. It would work a total of 3 times because -2 is a repeated root 3 times.
The only caution is when you're dealing with floating point numbers. This example worked out to all integers, but many solutions have floating point numbers which may cause a small remainder so you have to watch out for that. If the remainder is small be aware it may still be a repeated root. You can check by doing the math:
(m+1)*(m+2)*(m+2)*(m+2)*(m+4)
only with the floating point numbers, to see if it works out.
You might also run into this problem illustrated with this equation:
m^3+7.14159265*m^2+16.566371*m+12.566371=0
which gives solutions on the TI:
-1.99941, -2.00059, -3.14159
where we can see that the first two are almost the same, and the last one is almost equal to -pi. You have to judge whether or not the first two really are the same and represent a repeated root of say -2, while the last one is very close to -pi and may be the right solution if the original equation was given in more precision.