Question 151164
{{{(2x^2-x-9)/((x+1)(x^2+5))=A/(x+1)+(Bx+C)/(x^2+5)}}} Start with the given equation.



{{{(cross((x+1)(x^2+5)))((2x^2-x-9)/(cross((x+1)(x^2+5))))=(cross((x+1))(x^2+5))(A/cross((x+1)))+((x+1)cross((x^2+5)))((Bx+C)/cross((x^2+5)))}}} Multiply every term by the LCD {{{(x+1)(x^2+5)}}}. This will cancel out the denominators.



{{{2x^2-x-9=A(x^2+5)+(x+1)(Bx+C)}}} Multiply and simplify.



{{{2x^2-x-9=A(x^2+5)+Bx^2+Cx+Bx+C}}} FOIL



{{{2x^2-x-9=Ax^2+5A+Bx^2+Cx+Bx+C}}} Distribute




Take note that the {{{x^2}}} terms on the right side are {{{Ax^2}}} and {{{Bx^2}}}. So this means that {{{2x^2=Ax^2+Bx^2}}} and {{{2=A+B}}}. So the first equation is {{{A+B=2}}}



Also, notice how the {{{x}}} terms on the right side are {{{Bx}}} and {{{Cx}}}. So this means that {{{-x=Bx+Cx}}} and {{{2=B+C}}}. So the second equation is {{{B+C=-1}}}



Finally, notice how the constant terms on the right side are {{{5A}}} and {{{C}}}. So this means that {{{-9=5A+C}}}. So the third equation is {{{5A+C=-9}}}


{{{A+B=2}}} Start with the first equation.



{{{A=2-B}}} Subtract B from both sides.



{{{5A+C=-9}}} Move onto the third equation



{{{5(2-B)+C=-9}}} Plug in {{{A=2-B}}}



{{{10-5B+C=-9}}} Distribute



{{{-5B+C=-19}}} Subtract 10 from both sides.



So we now have the system of equations 


{{{system(B+C=-1,-5B+C=-19)}}}



Use any method to solve the system, to get the solutions:


{{{B=3}}} and {{{C=-4}}}



{{{A=2-B}}} Start with the given equation.



{{{A=2-3}}} Plug in {{{B=3}}}



{{{A=-1}}} Subtract





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Answer:


So the answers are


{{{A=-1}}}, {{{B=3}}}, and {{{C=-4}}}



Which means that 



{{{(2x^2-x-9)/((x+1)(x^2+5))=(-1)/(x+1)+(3x-4)/(x^2+5)}}}



Note: to check your answer, graph the left side and the right side. The two functions should be the same graph.