(algebra question) How do you factor and solve the following...?

Question

3x^2 + 5x - 12 = 0

5x^2 - 22x + 8 = 0

y^2 + 13y - 48 = 0

16x^2 - 25 = 0

2b^2 + 9b + 4 = 0


please and thank you

Answer 1

1. (3x - 4)(x + 3) = 0
3x - 4 = 0
3x = 4
x = 4/3
and
x + 3 = 0
x = -3

2. (5x - 2)(x - 4) = 0
5x + 2 = 0
5x = -2
x = -5/2
and x - 4 = 0
x = 4

3. (y + 16)(y - 3) = 0
y + 16 = 0
y = -16
and
y - 3 = 0
y = 3

4. Perfect square -> (4x+5)(4x-5) = 0
4x + 5 = 0
4x = -5
x = -4/5
and
4x - 5 = 0
4x = 5
x = 4/5

5. (2b + 1)(b + 4) = 0
2b + 1 = 0
2b = -1
b = -1/2
and
b + 4 = 0
b = -4

Answer 2

Mike gave you the first one.
There is a way to do these, but it still boils down to some trial and error. For example in problem 2, the coefficients of x and the constants must be such that the product of the coefficients is 5 and the sums of the coefficient-constant cross products is -22, and the product of the constants is 8. The numbers which do this are
(5x-2)(x-4)= 0 and x=4 or 2/5
I'll leave the others for you to do.
Number 4 is a difference of squares so the factors are (4x-5) and (4x+5), and x = +/- 5/4

Answer 3

to factor you take the first term times the last term and then find two multiples of that answer that equal the middle term

for example, the first problem...
3x-12=-36
factors: 9 and -4
so... 3x^2+9x-4x-12=0
3x (x+3) -4(x+3)=0
make the terms outside and inside the parenthesis factors...
(3x-4)(x+3)=0
then set each equal to zero to solve...
3x-4=0
x+3=0
x=4/3, x=-3

Answer 4

use foil

(3x-4)(x+3)=0 x=-3, x=4/3
...

Answer 5

quadratic equation