x2 + 5x + 3 = 9
x2 + 5x - 6 = 0
(x + 6)(x - 1) = 0
x = -6 or x = 1
x2+4x-9 = 5x+3 x2+4x-5x-9-3 = 0 x2-x-12 = 0 (x+3)(x-4) = 0 x = -3 or x = 4
x2 + 3x - 7 = 5x + 8; thus, x2 + 3x - 7 - 5x - 8 = x2 - 2x - 15 = (x - 5)(x + 3) = 0. Therefore the solution is x = 5 or -3.
x2 + 5x + 6 = (x + 2)(x + 3)
You don't factor an equation. You factor an expression.So we'll forget about the "equals 0" for a second, while we factor "x2 plus 5x plus 6".x2 + 5x + 6 = (x + 3) (x + 2).Now, if you want this expression to be equal to zero, that can only happen when 'x' is -2 or -3.
5x plus 3 does equal 3 plus 5x; commutative law.
x2 + 4x - 9 = 5x + 3 ∴ x2 - x - 12 = 0 ∴ (x + 3)(x - 4) = 0 ∴ x ∈ {-3, 4}
x2+4x-9 = 5x+3 x2+4x-5x-9-3 = 0 x2-x-12 = 0 (x+3)(x-4) = 0 x = -3 or x = 4
x2 + 3x - 7 = 5x + 8; thus, x2 + 3x - 7 - 5x - 8 = x2 - 2x - 15 = (x - 5)(x + 3) = 0. Therefore the solution is x = 5 or -3.
x2 + 5x + 6 = (x + 2)(x + 3)
You don't factor an equation. You factor an expression.So we'll forget about the "equals 0" for a second, while we factor "x2 plus 5x plus 6".x2 + 5x + 6 = (x + 3) (x + 2).Now, if you want this expression to be equal to zero, that can only happen when 'x' is -2 or -3.
5x plus 3 does equal 3 plus 5x; commutative law.
x2 - 5x + 6 = (x - 2)(x - 3)
x2+5x+6 = (x+2)(x+3)
x2+5x+6 = (x+2)(x+3) when factored because 2 and 3 are factors of 6
33
=(x+3)(x+2)
I suggest you use the quadratic formula. In this case, a = 1, b = 5, c = 3.