Question 1171058

Perform partial fraction decomposition for the expression 

{{{(8+x+6x^2-12x^3)/ ((3x^2+4)(x^2+7))}}}

{{{(-12x^3+6x^2+x+8)/ ((3x^2+4)(x^2+7))}}}

Factor the numerator and denominator:

{{{(-12x^3+6x^2+x+8)/ ((3x^2+4)(x^2+7))=(3(-4x^3+2x^2+x/3+8/3))/ (3(x^2+4/3)(x^2+7))}}}

The form of the partial fraction decomposition is

{{{(-4x^3+2x^2+x/3+8/3)/ ((x^2+4/3)(x^2+7))=(Ax+B)/(x^2+7)+(Cx+D)/(x^2+4/3)}}}

Write the right-hand side as a single fraction:


{{{(-4x^3+2x^2+x/3+8/3)/ ((x^2+4/3)(x^2+7))=((Ax+B)(x^2+4/3)+(Cx+D)(x^2+7))/((x^2+4/3)(x^2+7))}}}

The denominators are equal, so we require the equality of the numerators:

{{{-4x^3+2x^2+x/3+8/3=(Ax+B)(x^2+4/3)+(Cx+D)(x^2+7)}}}

Expand the right-hand side:

{{{-4x^3+2x^2+x/3+8/3=3x^3A+3x^3C+3x^2B+3x^2D+4xA+21xC+4B+21D}}}

Collect up the like terms:

{{{-4x^3+2x^2+x/3+8/3=(3A+3C)x^3+(3B+3D)x^2+(4A+21C)x+4B+21D}}}

The coefficients near the like terms should be equal, so the following system is obtained:

{{{3A+3C=-4}}}
{{{3B+3D=2}}}
{{{4A+21C=1/3}}}
{{{4B+21D=8/3}}}

Solving it (using calculator) we get that {{{A=-5/3}}}, {{{B=2/3}}},{{{ C=1/3}}}, {{{D=0}}}

Therefore,

{{{(-4x^3+2x^2+x/3+8/3)/ ((x^2+4/3)(x^2+7))=(Ax+B)/(x^2+7)+(Cx+D)/(x^2+4/3)}}}

{{{(-4x^3+2x^2+x/3+8/3)/ ((x^2+4/3)(x^2+7))=((-5/3)x+2/3)/(x^2+7)+((1/3)x+0)/(x^2+4/3)}}}
{{{(-4x^3+2x^2+x/3+8/3)/ ((x^2+4/3)(x^2+7))=((-5/3)x+2/3)/(x^2+7)+(x/3)/((3x^2+4)/3)}}}
{{{(-4x^3+2x^2+x/3+8/3)/ ((x^2+4/3)(x^2+7))=(2/3-5x/3)/(x^2+7)+(x/cross(3))/((3x^2+4)/cross(3))}}} ...simplfy

Answer:

{{{(-4x^3+2x^2+x/3+8/3)/ ((x^2+4/3)(x^2+7))=(2/3-5x/3)/(x^2+7)+x/(3x^2+4)}}}