Answer:
[tex]$-\frac{1}{4} (-2x-9+\sqrt{65})(2x+9+\sqrt{65})$[/tex]
Step-by-step explanation:
We have a second-degree polynomial:
I will find the roots (y = 0) completing the square:
[tex]$x^2+9x+\frac{81}{4}-\frac{81}{4}+4=0$[/tex]
[tex]$\left(x+\frac{9}{2} \right)^2 -\frac{81}{4} +4 =0$[/tex]
[tex]$\left(x+\frac{9}{2} \right)^2 -\frac{81}{4} +\frac{16}{4} =0$[/tex]
[tex]$\left(x+\frac{9}{2} \right)^2 -\frac{65}{4} =0$[/tex]
[tex]$\left(x+\frac{9}{2} \right)^2 =\frac{65}{4} $[/tex]
[tex]$\left(x+\frac{9}{2} \right)^2 =\pm \sqrt{\frac{65}{4} } $[/tex]
[tex]$x =-\frac{9}{2} \pm \frac{\sqrt{65} }{2} $[/tex]
[tex]$x =\frac{-9 \pm \sqrt{65} }{2} $[/tex]
[tex]$\boxed{x_1 =\frac{-9 + \sqrt{65} }{2}} $[/tex]
[tex]$\boxed{x_2 =\frac{-9 - \sqrt{65} }{2}} $[/tex]
