Question 151438
{{{(11x^2-x+8)/((x+1)(x^2+4))=A/(x+1)+(Bx+C)/(x^2+4)}}} Start with the given equation.



{{{(cross((x+1)(x^2+4)))((11x^2-x+8)/(cross((x+1)(x^2+4))))=(cross((x+1))(x^2+4))(A/cross((x+1)))+((x+1)cross((x^2+4)))((Bx+C)/cross((x^2+4)))}}} Multiply <b>every</b> term by the LCD {{{(x+1)(x^2+4)}}} to clear the fractions.



{{{11x^2-x+8=(x^2+4)(A)+(x+1)(Bx+C)}}} Simplify



{{{11x^2-x+8=A(x^2+4)+(x+1)(Bx+C)}}} Rearrange the terms



{{{11x^2-x+8=A(x^2+4)+Bx^2+Cx+Bx+C}}} FOIL



{{{11x^2-x+8=Ax^2+4A+Bx^2+Cx+Bx+C}}} Distribute






Notice how the {{{x^2}}} terms on the right side are {{{Ax^2}}} and {{{Bx^2}}}. So this means that {{{Ax^2+Bx^2=11x^2}}} and {{{A+B=11}}}



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


Finally, the constant terms on the right side are {{{4A}}} and {{{C}}}. So this means that {{{4A+C=8}}} 




{{{A+B=11}}} Start with the first equation.



{{{B=11-A}}} Subtract A from both sides.



{{{B+C=-1}}} Move onto the second equation



{{{11-A+C=-1}}} Plug in {{{B=11-A}}}



{{{-A+C=-12}}} Subtract 11 from both sides.



{{{-4A+4C=-48}}} Multiply both sides by 4.



Add this equation to equation #3


{{{-4A+4C=-48}}}
+{{{4A+C=8}}}
--------------
{{{0A+5C=-40}}}



{{{5C=-40}}} Simplify



{{{C=-8}}} Divide both sides by 5.



{{{4A+C=8}}} Go back to the third equation



{{{4A-8=8}}} Plug in {{{C=-8}}}



{{{4A=16}}} Add 8 to both sides.



{{{A=4}}} Divide both sides by 4.



{{{B=11-A}}} Move onto the first isolated equation



{{{B=11-4}}} Plug in {{{A=4}}}



{{{B=7}}} Subtract



====================================================


So the value of the constants are 


{{{A=4}}}, {{{B=7}}}, and {{{C=-8}}}



So this means that the equation is {{{(11x^2-x+8)/((x+1)(x^2+4))=(4)/(x+1)+(7x-8)/(x^2+4)}}}