Question 1171058
<br>
First a comment about the sloppy presentation of the expression....<br>
Here is how the program on this website interprets the expression as you show it:<br>
{{{8+x+6x^2-12x^3/ (3x^2+4)(x^2+7)}}}<br>
That is not the standard interpretation of your expression.  By standard rules for order of operations, your expression would be interpreted as<br>
{{{8+x+6x^2-(12x^3/ (3x^2+4))*(x^2+7)}}}<br>
Clearly neither of those expressions is what you meant to show; a partial fraction decomposition does not make sense for either of those expressions.<br>
If you are working on a problem like this, then your knowledge of mathematics is surely advanced enough to know that parentheses are sometimes required to make an expression say what you intended it to say....<br>
The correct form of your expression is (8+x+6x^2-12x^3)/ ((3x^2+4)(x^2+7)):<br>
{{{(8+x+6x^2-12x^3)/ ((3x^2+4)(x^2+7))}}}<br>
The decomposition will be<br>
{{{(Ax+B)/(3x^2+4)+(Cx+D)/(x^2+7)}}}<br>
So we have<br>
{{{(8+x+6x^2-12x^3)/ ((3x^2+4)(x^2+7)) = (Ax+B)/(3x^2+4)+(Cx+D)/(x^2+7)}}}<br>
Combine the two fractions on the right using a common denominator:<br>
{{{((Ax+B)(x^2+7)+(Cx+D)(3x^2+4))/((3x^2+4)(x^2+7))}}}<br>
Now the denominators on both sides of the equation are the same, so the numerators must be the same.  Expand the numerator on the right and equate coefficients to get systems of equations that you can solve to find A, B, C, and D.<br>
{{{-12x^3+6x^2+x+8 = (Ax^3+Bx^2+7Ax+7B)+(3Cx^3+3Dx^2+4Cx+4D)}}}
{{{-12x^3+6x^2+x+8 = (A+3C)x^3+(B+3D)x^2+(7A+4C)x+(7B+4D)}}}<br>
{{{A+3C = -12}}}
{{{B+3D = 6}}}
{{{7A+4C = 1}}}
{{{7B+4D = 8}}}<br>
Solving those equations gives (A,B,C,D) = (3,0,-5,2), and the decomposition is<br>
{{{3x/(3x^2+4)+(-5x+2)/(x^2+7)}}}<br>