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