Express the function as the sum of a power series by first using partial fractions. f(x) = 8 x2 − 4x − 12 f(x) = ∞ n = 0 find the interval of convergence. (enter your answer using interval notation.)

I'm guessing the function is[tex]f(x)=\dfrac8{x^2-4x-12}=\dfrac8{(x-6)(x+2)}[/tex]which, split into partial fractions, is equivalent to[tex]\dfrac1{x-6}-\dfrac1{x+2}[/tex]Recall that for [tex]|x|<1[/tex] we have[tex]\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n[/tex]With some rearranging, we find[tex]\dfrac1{x-6}=-\dfrac16\dfrac1{1-\frac x6}=\displaystyle-\frac16\sum_{n=0}^\infty\left(\frac x6\right)^n[/tex]valid for [tex]\left|\dfrac x6\right|<1\[/tex], or [tex]|x|<6[/tex], and[tex]\dfrac1{x+2}=\dfrac12\dfrac1{1-\left(-\frac x2\right)}=\displaystyle\frac12\sum_{n=0}^\infty\left(-\frac x2\right)^n[/tex]valid for [tex]\left|-\dfrac x2\right|<1[/tex], or [tex]|x|<2[/tex].So we have[tex]f(x)=\displaystyle-\frac16\sum_{n=0}^\infty\left(\frac x6\right)^n-\frac12\sum_{n=0}^\infty\left(-\frac x2\right)^n[/tex][tex]f(x)=\displaystyle-\sum_{n=0}^\infty\left(\frac{x^n}{6^{n+1}}+\frac{(-x)^n}{2^{n+1}}\right)[/tex][tex]f(x)=\displaystyle-\sum_{n=0}^\infty\frac{1+3(-3)^n}{6^{n+1}}x^n[/tex]Taken together, the power series for [tex]f(x)[/tex] can only converge for [tex]|x|<2[/tex], or [tex]-2<x<2[/tex].