how do u solve these 2 problems???

Question

1. x^2 + 3x - 31 = -3
2. x^2 2x - 19 = -4

using factoring , please!!!thanx

Answer 1

1. Add 3 to both sides to set equal to zero, then factor to get something that adds to three and multiplies to 28---7 and -4
(x-4)(x+7) = 0
x=4, 7

2. Add 4 to both sides to set equal to zero, then factor to get something that adds to 2 and multiplies to -15---5 and -3
(x+5)(x-3) = 0
x = -5, 3

Answer 2

Make one side of each of the equations zero first.
x^2+3x-31=-3 Add three to both sides
x^2+3x-28=0

x^2+2x-19=-4 Add four to both sides
x^2+2x-15=0

Now factor each equation
x^2+3x-28=(x-4)(x+7)
x^2+2x-15=(x+5)(x-3)

Now set all of the factors equal to zero and solve
First one: x=0 @ 4 and -7
Second one: x=0 @ -5 and 3

Answer 3

1. First, add three to both sides:
x^2 + 3x - 28=0
Then factor into (x-4)(x+7)=0
Then from the Zero Product Property, which states that if ab=0, then a or b or both is zero,
x-4=0
x+7=0
Solve both of these problems.
x=4, x=-7

2. x^2 + 2x - 19 = -4 (You missed a plus sign, but it may have been minus. I'll just assume it's a plus sign.)
x^2 + 2x -15=0
(x-3)(x+5)=0
x=3, x=-5

Answer 4

x^2 + 3x - 32 = -3
x^2 + 3x - 28 = 0
(x + 7)(x - 4)
x = -7 or 4

---------------------------

x^2 + 2x - 19 = -4
x^2 + 2x - 15 = 0
(x + 5)(x - 3) = 0
x = -5 or 3

unless you meant

x^2 - 2x - 19 = -4
x^2 - 2x - 15 = 0
(x - 5)(x + 3) = 0
x = 5 or -3

Answer 5

one of them gotta be right...

Answer 6

huh?