Question

1" title="0= \frac{3 x^{2}-18x+24 }{ x^{2} -2x-8} " alt="0= \frac{3 x^{2}-18x+24 }{ x^{2} -2x-8} " align="absmiddle" class="latex-formula">
Solve for x
Please show your work!!!
Thanks in advance :)

Answer 1

$D:x^2-2x-8\not=0\\ D:x^2-4x+2x-8\not=0\\ D:x(x-4)+2(x-4)\not=0\\ D:(x+2)(x-4)\not=0\\ D:x\not =-2 \wedge x\not =4\\\\ \frac{3x^2-18x+24}{x^2-2x-8}=0\\ \frac{3(x^2-6x+8)}{(x+2)(x-4)}=0\\ \frac{3(x^2-4x-2x+8)}{(x+2)(x-4)}=0\\ \frac{3(x(x-4)-2(x-4))}{(x+2)(x-4)}=0\\ \frac{3(x-2)(x-4)}{(x+2)(x-4)}=0\\ \frac{3(x-2)(x-4)}{(x+2)(x-4)}=0\\ \frac{3(x-2)}{(x+2)}=0|\cdot(x+2)\\ 3(x-2)=0|:3\\ x-2=0\\ \boxed{x=2}$