Question 442453
Notice that there is a "b" common in every term of this trinomial.

Thus we can factor out a b. So {{{b^3-5b^2-14b = b(b^2-5b-14)}}}

Now we must factor that quadratic.  What multiplies to -14 and adds to -5?

Here are the factors of 14.. plus or minus

1  14
2  7

Ah ha! 2-7 = -5.

So our factorization would be (b+2)(b-7)

Going back, that means our final factorization is {{{b*(b+2)*(b-7)}}}