To answer this question, we can proceed as follows:
1. We can multiply by the least common multiple to both sides of the next equation:
[tex]\begin{gathered} \frac{-8}{x-16}+\frac{-1}{x+17}=0 \\ \operatorname{lcm}(x-16,x+7)=(x-16)(x+17) \\ \end{gathered}[/tex]2. Then we have:
[tex](x-16)(x+17)(\frac{-8}{x-16}+\frac{-1}{x+17})=(x-16)(x+17)\cdot0[/tex][tex](x-16)(x+17)\frac{-8}{x-16}+(x-16)(x+17)\frac{-1}{x+17})=0[/tex][tex]\frac{(x-16)}{(x-16)}\cdot(x+17)\cdot(-8)+\frac{(x+17)}{(x+17)}\cdot(x-16)(-1)=0[/tex][tex]\frac{a}{a}=1,\frac{(x-16)}{(x-16)}=1,\frac{(x+17)}{(x+17)}=1[/tex]Then we have:
[tex]\begin{gathered} (x+17)(-8)+(x-16)(-1)=0 \\ -8(x)+17(-8)+(x)(-1)+(-16)(-1)=0 \\ -8x-136-x+16=0 \\ -8x-x-136+16=0 \end{gathered}[/tex]If we add the like terms, we have:
[tex]-9x-120=0[/tex]3. Adding 120 to both sides of the equation, and then dividing both sides of the equation by -9:
[tex]\begin{gathered} -9x-120+120=120 \\ -9x=120 \\ \frac{-9}{-9}x=\frac{120}{-9} \\ x=-\frac{\frac{120}{3}}{\frac{9}{3}}=-\frac{40}{3} \\ x=-\frac{40}{3} \end{gathered}[/tex]In summary, therefore, the value of x that makes f(x) = 0 is:
[tex]x=-\frac{40}{3}[/tex]