solving quadratic equation by factoring?

Question

1.x2+x-6=0
2.2x2+5x+2=0
3.4x2-5x+2=0
4.2x2+5x+2=0
5.x2-x-2=0

Answer 1

1. x+3, x-2
2. 2x +1, x+2
3. 2x+2, 2x+1
4. same as 3
5. x-2, x + 1

Answer 2

1) x^2+x-6=0
(x-2)(x+3) = 0
x = 2 or x = -3

2) 2x^2+5x+2=0
(2x+1)(x+2) = 0
2x+1=0 or x+2=0
2x = -1 or x = -2
x= -1/2 or x = -2

3) 4x^2-5x+2=0
By using the formula x={-B+or-[sqrt (B^2-4AC)]}/2A where Ax^2+Bx+C=0.
x={-(-5)+or-[sqrt ((-5)^2-4(4)(2))]}/2(4)
x=[5 +or-sqrt(25-32)]/8
x=[5 +or-sqrt(-7)]/8

x is insolvable becos sqrt(-7) is undefined.

4) 2x^2+5x+2=0
Look at Q2...same question

5) x^2-x-2=0
(x-2)(x+1)=0
x-2 = 0 or x+1=0
x=2 or x=-1

Answer 3

1. (x+3)(x-2)
2. (2x+1)(x+2)
3. (4x-2)(x-1)
4. (2x+1)(x+2)
5. (x-2)(x+1)