Partial Fraction Decomposition of $\frac{7x^2-3x+45}{x^3+9x}$

Question

Consider the fraction $$\frac{7x^2-3x+45}{x^3+9x}$$

So after factoring the denominator and putting them under $A$,$B$ and, $C$

I get

$$\frac{A(x^2+9)+Bx+C(x)}{(x)(x^2+9)}$$

Now that the denominator is the same as the original I get

$$7x^2-3x+45=A(x^2+9)+Bx+C(x)$$ After expanding the right side and grouping like terms I get

$$7x^2-3x+45=x^2(A+B)+Cx+9A$$ I set $x=0$ and get $A=5$

Seeing that the forms are the same means that

$(A+B)=7$

$C=-3$

$9A=45$

but I don't know how to find $B$ and $C$

Answer 1

$C=-3$

$9A=45$ $\implies$ $A=5$

$A+B=7$ and $A=5$ $\implies$ $B=2$.

Alternatively,

Put $x=0$ into $\displaystyle 7x^2-3x+45=A(x^2+9)+(Bx+C)(x)$, we have $A=5$

To find $B$ and $C$, we can just substitute two other numbers into $x$ (e.g. $x=1$ and $x=-1$).

We can even do differentiation and substitution, although it is not really necessary.

Differentiating both sides with respect to $x$,

$$14x-3=2(A+B)x+C$$

Put $x=0$ and obtain $C=-3$

Differentiating again,

$$14=2(A+B)=2(5+B) \implies B=2$$

You missed some brackets:

$\displaystyle \frac{7x^2-3x+45}{x^3+9x}=\frac{A}{x}+\frac{Bx+C}{x^2+9}=\frac{A(x^2+9)+(Bx+C)(x)}{x(x^2+9)}$

$\displaystyle 7x^2-3x+45=A(x^2+9)+(Bx+C)(x)$

Answer 2

Using complex ,

$$f(x)=\frac Ax+\frac{Bx+C}{x^2+9}$$

Let $$g(x)=xf(x)$$ and $$h(x)=(x^2+9)f(x)$$ then

$$A=g(0)=\frac{45}{9}=5$$

and

$$3iB+C=h(3i)=\frac{-63-9i+45}{3i}$$ $$=21i-3-15i=-3+6i$$ thus $$B=2 \text{ and } C=-3$$