Question 37169
(1) Here is the first
m^3+6m^2+9m=0  , This polynomial of degree 3 is divisible by m, therefore we divede by m and get:

m(m^2+6m+9)=0 Now we have to solve the polynomial of degree two, and the solution is m=-3, and is unique. Finally we get

m(m+3)(m+3) As the factorization of m^3 + 6m^2 + 9m 

(2) Here is the second
b^2-6b-10=0, First have to solve the equation, find the roots, we can do this using the general formula or using more advance methods. Using the formula we get as a result b=3+sqrt(19), and b=3-sqrt(19). Then the factors will be

(b-(3-sqrt(19))(b-(3+(sqrt(19)) 

{{{b^2-6b-10=(b-(3-sqrt(19)))(b-(3+sqrt(19)))}}}