Question

Are (3c + 7) and (7c + 3) the binomial factors of 21c2 + 58c + 21?

Answer 1

(3c + 7)(7c + 3)
21c^2 + 49c + 9c + 21
21c^2 + 58c + 21
Yes, because when multiplied together they make 21c^2 + 58c + 21
Hope this helps!

Answer 2

Hello :),
=(3c+7)(7c+3)
=21c²+9c+49c+21
=21c²+58c+21