How do you factor the problems below? Im confused.?

Question

Please help me to understand how to factor these because im confused...or are they prime?
5c^2-4d^2


36j^2-49m



-4c^2+25


4g^2-81h^2

thank you

Answer 1

5c^2-4d^2= (2.24c - 2d)(2.24c + 2d)


36j^2-49m = (6j -7sqrt(m))(6j + 7 sqrt(m))



-4c^2+25 = (5 - 2c)(5 + 2c)



4g^2-81h^2 = (2g - 9h)(2g + 9h)

Just remember to do squart root of the first number and the second number then change the signs of the two parts.

Answer 2

#1. I don't know if this is what you call prime, but it can't be factored further into integer coefficients.
#2 This is a sum and difference of perfect squares of a sort, so it factors
6j-7sqrt(m) times 6j+7sqrt(m)
#3 Other than the sneaky sign switch, this is also ths sum and difference of perfect squres as is the next one. (a^2 - b^2) = (a-b)(a+b). Just pick your a's and b's.

Answer 3

[sqrt(5)c + 2d] [sqrt(5)c - 2d]

(6j+7m)(6j-7m) if the problem is 36j^2-49m^2

25 - 4c^2 = (5+2c)(5-2c)

4g^2-81h^2 = (2g + 9h)(2g - 9h)