Question 1055414
{{{ (-x^2+3x+8)/(x^3+4x^2+4x) =(-x^2+3x+8)/(x(x+2)^2) }}}
{{{(-x^2+3x+8)/(x(x+2)^2)=A/x+B/(x+2)+C/(x+2)^2 }}}
{{{-x^2+3x+8=A(x+2)^2+Bx(x+2)+Cx}}}
{{{-x^2+3x+8=A(x^2+4x+4)^2+B(x^2+2x)+Cx}}}
So,
{{{A+B=-1}}}
{{{4A+2B+C=3}}}
{{{4A=8}}}
Then,
{{{A=2}}}
and
{{{2+B=-1}}}
{{{B=-3}}}
and
{{{4(2)+2(-3)+C=3}}}
{{{8-6+C=3}}}
{{{C=1}}}
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{{{ (-x^2+3x+8)/(x^3+4x^2+4x) =2/x-3/(x+2)+1/(x+2)^2 }}}