Me i'm a desperate sobbing mess cause of this class. solve by factoring 1) 2z^2-36=0 2) c^2=5c 3) p^2-4p=21 4) 2w^2-11w= -12 write the equation in standard form. then solve. 5) 4q^2+3q^2-4q=18

Don't sob! I gotchu!

2. c^2=5c aka c^2-5c+0=0 right?
now we need a number on the left which will multiply to give c^2, we will get that by c*c good?
now to the right side, we need numbers that multiply to give 0 and sum to -5, we'll get that by -5*0=0 and -5+0=-5 good?
now you get the factor (c-5)!

now this way, while factoring an equation that has a coefficient of more than 1, we multiply that coefficient to the constant (in the first question, 2 will be multiplied to 36 giving 72) now we need numbers that multiply to give 72 and sum to 0, since there isn't a number like that, we will have to use the quadratic formula. get it?

(also if the alphabets get too confusing, you can always use x, it doesn't matter!)

31 is prime. and it’s prime factorization is just 31.