SOLUTION: Square the binomial (7a+4y)^2 (7a+4y)(7a+4y) F (7a)(7a)=49a^2 O (7a)(4y)= 28ay I (4y)(7a) = 28ay L (4y) (4y) = 16y^2 49a^2+56ay+16y^2

Algebra ->  Distributive-associative-commutative-properties -> SOLUTION: Square the binomial (7a+4y)^2 (7a+4y)(7a+4y) F (7a)(7a)=49a^2 O (7a)(4y)= 28ay I (4y)(7a) = 28ay L (4y) (4y) = 16y^2 49a^2+56ay+16y^2      Log On


   

Question 70722: Square the binomial
(7a+4y)^2
(7a+4y)(7a+4y)
F (7a)(7a)=49a^2
O (7a)(4y)= 28ay
I (4y)(7a) = 28ay
L (4y) (4y) = 16y^2
49a^2+56ay+16y^2
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Square the binomial
(7a+4y)^2
(7a+4y)(7a+4y)
F (7a)(7a)=49a^2
O (7a)(4y)= 28ay
I (4y)(7a) = 28ay
L (4y) (4y) = 16y^2
49a^2+56ay+16y^2
Correct
Cheers,
Stan H.