Answer to Question #208429 in Calculus for Areeba

Question

 Reduce the following fraction into partial fractions: 3x^ 2 + 7x + 28/x(x^ 2 + x + 7)

Answer 1

$$\dfrac{3x^2+7x+28}{x(x^2+x+7)}=\dfrac{A}{x}+\dfrac{B}{x^2+x+7}+\dfrac{Cx}{x^2+x+7}$$

$$=\dfrac{Ax^2+Ax+7A+Bx+Cx^2}{x(x^2+x+7)}$$

$x^2: A+C=3$

$x^1: A+B=7$

$x^0: 7A=28$

$A=4$

$B=3$

$C=-1$

$$\dfrac{3x^2+7x+28}{x(x^2+x+7)}=\dfrac{4}{x}+\dfrac{3}{x^2+x+7}-\dfrac{x}{x^2+x+7}$$