The partial fraction decomposition of LaTeX: \frac{x-9}{x^2-3x-18} x − 9 x 2 − 3 x − 18 is LaTeX: \frac{A}{x-6}+\frac{B}{x+3} A x − 6 + B x + 3 . Find the numbers LaTeX: A\: A and LaTeX: B B . Then, find the sum LaTeX: A+B A + B , which is a whole number. Enter that whole number as your answer.

Accepted Solution

Not entirely sure what the question is supposed to say, so here's my best guess.First, find the partial fraction decomposition of[tex]\dfrac{x-9}{x^2-3x-18}[/tex]This is equal to[tex]\dfrac{x-9}{(x-6)(x+3)}=\dfrac a{x-6}+\dfrac b{x+3}[/tex]Multiply both sides by [tex](x-6)(x+3)[/tex], so that[tex]x-9=a(x+3)+b(x-6)[/tex]Notice that if [tex]x=6[/tex], the term involving [tex]b[/tex] vanishes, so that[tex]6-9=a(6+3)\implies a=-\dfrac13[/tex]Then if [tex]x=-3[/tex], the term with [tex]a[/tex] vanishes and we get[tex]-3-9=b(-3-6)\implies b=\dfrac43[/tex]So we have[tex]\dfrac{x-9}{x^2-3x-18}=-\dfrac1{3(x-6)}+\dfrac4{3(x+3)}[/tex]I think the final answer is supposed to be [tex]a+b[/tex], so you end up with 1.