Factor the polynomial completetly 21b2-40b-21

First we note symmetry in the expression's coefficients.
We also note that 7*3=21, and 7+3=10.

From the rational roots theorem, we are tempted to try with 3 and 7 as coefficients of the factors.

Try
(7b+3)(3b+7)=21b^2+(49+9)b+21
By switching the sign of 3b+7 to 3b-7, we get the signs right, to check:
(7b+3)(3b-7)=21b^2+(9-49)b-21=21b^2-40b-21  right!

So
(7b+3)(3b-7)=21b^2-40b-21