Question

Consider the decomposition of

6x3+x−3/x2−2x+1

into partial fractions

Answer 1

Use long division to find the quotient and remainder of $\frac{6x^3+x-3}{x^2-2x+1}$ and then rewrite it as the quotient plus the remainder over the denominator:

Now that we have that $\frac{6x^3+x-3}{x^2-2x+1}=(6x+12)+\frac{19x+15}{x^2-2x+1}$. We have only to find the partial fraction decomposition of $\frac{19x-15}{x^2-2x+1}$.

Express as a ratio of expanded polynomials in lowest terms: $\frac{19x-15}{x^2-2x+1}=\frac{19x-15}{(x-1)^2}$.

The partial fraction expansion is of the form: $\frac{19x-15}{(x-1)^2}=\frac{a}{x-1}+\frac{b}{(x-1)^2}$.

Multiply both sides by $(x-1)^2$ and simplify: $19x-15=a(x-1)+b$.

Expand and collect in terms of powers of $x$: $19x-15=-a+bx+c$.

Equate coefficients on both sides, yielding 2 equation in 2 unknows:

$\begin{cases} -15=b-a\\19=a \end{cases}$

The solution: $a=19$, $b=4$.

Therefore: $\frac{19x-15}{x^2-2x+1}=\frac{4}{(x-1)^2}+\frac{19}{x-1}$.

Compiling our results, we have that $\frac{6x^3+x-3}{x^2-2x+1}=12+6x+\frac{4}{(x-1)^2}+\frac{19}{x-1}$.