[tex]a {x}^{2} + bx + c = a(x - x_{1})(x - x_{2})[/tex]

По теореме Виета:

[tex]{x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c[/tex]

1) В

[tex] - {x}^{2} + 8x + 9 = \\ {x}^{2} - 8x - 9 = 0 \\ x_{1} + x_{2} = 8 \\ x_{1} x_{2} = - 9 \\ x_{1} =9 \\ x_{2} = - 1 \\ - {x}^{2} + 8x + 9 = - (x + 1)(x - 9)[/tex]

2) А

[tex] {x}^{2} - 8x - 9 = (x + 1)(x - 9)[/tex]

3) Д

[tex] {x}^{2} + 5x + 6 = 0 \\ x_{1} + x_{2} = - 5 \\ x_{1} x_{2} = 6 \\ x_{1} = - 3\\ x_{2} = - 2 \\ {x}^{2} + 5x + 6 = (x + 2)(x + 3)[/tex]

4) Б

[tex] - x + {x}^{2} - 6 = 0 \\ {x}^{2} - x - 6 = 0 \\ x_{1} + x_{2} = 1 \\ x_{1} x_{2} = - 6 \\ x_{1} = 3\\ x_{2} = - 2 \\ - x + {x}^{2} - 6 = (x + 2)(x - 3)[/tex]