How to determine h(x) in a polynomial partial fraction decompostion

Question

Im susposed to do a partial fractional division;

$$ \frac{-2x^2 + 8x - 9} {(x-1) (x-3)^2}$$ Now I used the formula and this is what I got;

$$\frac {A}{(x-1)} + \frac{B}{x-3} + \frac{C}{(x-3)^2}$$

Now I've put the 3 fractions together,and got this

$$ \frac{A(x-3)^2+B(x-3)(x-1) +C(x-1)} {(x-1) (x-3)} $$

Now I looked at the starting function looked what my x^2 ,x terms were on the right on the left, solved it and got this

$$f(x) = \frac{-2} {4(x+1)} + \frac{-5} {4(x-3)} + \frac {3} {2(x-1)^2} $$

Now after looking at the formula I'd realise that I'm missing my h(x) the whole part if you will ( not sure how it is said in english) Now the h(x) doesnt exist when m < n but in my case m = n so I should also be having that? But how do I get that part? the formula says simply h(x) + the A B and C parts.How do I calculate h(x)?

EDIT: To clarify what h(x) means and what I was exactly trying to figure out.

$$f(x) = h(x) + \frac{r(x)} {q(x)}$$

Where m = deg r and n = deg q

Answer 1

You made a mistake when you put three fractions together: $$ \frac{A(x-3)^2+B(x-1)(x-3)+C(x-1)}{(x-1)(x-3)^\color{red}{2}}. $$

Now you set $$ A(x-3)^2+B(x-1)(x-3)+C(x-1)=-2x^2+8x-9 $$ to find $A,B,C$.

• Set $x=1$: $4A=-2+8-9$.
• Set $x=3$: $2C=-18+24-9$.
• Set $x=2$: $A-B+C=-8+16-9$.

Now it should be very straightforward to go on.

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Notes.

If you write $$ f(x)=\frac{-2x^2+8x-9}{(x-1)(x-3)^2} $$ then you can find polynomials $h,p,q$ so that $$ f(x)=h(x)+\frac{p(x)}{q(x)} $$ where the $\deg p<\deg q$. This is the preliminary step for partial fraction decomposition.

But note that $$ \deg (-2x^2+8x-9)<\deg (x-1)(x-3)^2 $$ you have $h(x)=0$, $p(x)=-2x^2+8x-9$ and $q(x)=(x-1)(x-3)^2$.

Answer 2

Your approach is fine, but there is some mistake in your computations, since, in fact$$A=-\frac34,\ B=-\frac54\text{, and }C=-\frac32;$$therefore$$\frac{-2x^2+8x-9}{(x-1)(x-3)^2}=-\frac3{4(x-1)}-\frac5{4(x-3)}-\frac3{2(x-3)^2}.$$And, in your case, the degree of the numerator is $2$ and the degree of the denominator is $3$. Therefore, there is nothing else to be done.