factoring
posted by mary .
how do you factor
m^3+5m^2+3m9
it might be simple i am not seeing it..
please help so i can continue with the problem...thanks

Well, by inspection, I see that m=1 works, so divide m1 into the polynomial to see what is left.

okey i am confused. We are suppose to solve for what m is equal to...i thought we could factor by grouping...but didn't know exactly how..please explain

one factor is by inspection (m1)
Divide then m^3+5m^2+3m9 by (m1) and you will have the second factor, which I suspect can then be factored easily. 
i honestly don't see how u divid it and u get the answer...please explain in detail

You can also use the Rational Roots Theorem. If m = p/q is a solution, then p has to be a divisor of 9 and q a divisor of 1 (9 is the constant term and 1 is the coefficient of the highest power of m).
This means that all rational solutins must be divisors of 9.
So, you only have to try m = 1, m = 1, m= 3, m = 3, m = 9 and m = 9.
I besides m = 1, m = 3 is a solution. If you had found no other solution besides m = 1, then that would have menat that the other solutioins are not rational numbers. If we find one other, then that means that the remaining one must also be a rational number. But if only m = 1 and m = 3 are the possible rational solutions, this means that one of the roots is a double root.
In this case, you fiund that m = 3 is the double root, and the factorization is:
(m1)(m+3)^2
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