Answer:
[tex]-39x^{2} \sqrt{5}[/tex]
Step-by-step explanation:
Firstly, you have to know these:
[tex]a\sqrt{x} + b\sqrt{x} = (a+b)\sqrt{x} \\\sqrt{a.b^{2} } =b\sqrt{a} \\ c\sqrt{a.b^{2} } =(c.b)\sqrt{a} \\a\sqrt{x} - b\sqrt{x} = (a-b)\sqrt{x}[/tex]
About your question :
[tex]\sqrt{180} = \sqrt{6^{2}.5} = 6\sqrt{5}[/tex]
[tex]-8x^{2} . 6\sqrt{5} = -48x^{2}\sqrt{5}[/tex]
[tex]\sqrt{5x^{2}} = x\sqrt{5}[/tex]
[tex]9x\sqrt{5x^{2}} = 9x.x\sqrt{5} = 9x^{2} \sqrt{5}[/tex]
-48x²√5 + 9x²√5 = -39√5
Hope this helps ^-^
