Question

`0= (3 x^(2)-18x+24 )/( x^(2) -2x-8)`

Solve for x
Please show your work!!!
Thanks in advance :)

Answer 1

`D:x^2-2x-8\\ot=0\\ D:x^2-4x+2x-8\\ot=0\\ D:x(x-4)+2(x-4)\\ot=0\\ D:(x+2)(x-4)\\ot=0\\ D:x\\ot =-2 \wedge x\\ot =4\\\\ (3x^2-18x+24)/(x^2-2x-8)=0\\ (3(x^2-6x+8))/((x+2)(x-4))=0\\ (3(x^2-4x-2x+8))/((x+2)(x-4))=0\\ (3(x(x-4)-2(x-4)))/((x+2)(x-4))=0\\ (3(x-2)(x-4))/((x+2)(x-4))=0\\ (3(x-2)(x-4))/((x+2)(x-4))=0\\ (3(x-2))/((x+2))=0|\cdot(x+2)\\ 3(x-2)=0|:3\\ x-2=0\\ \boxed{x=2}`