Partial Fractions
Polynomial fractions of the form $\dfrac{ax+b}{cx^2+dx+e}$ can often be simplified by means of partial fractions. If the denominator can be factorised then the fraction can be split into the sum of simpler fractions.
Consider the fraction $\dfrac{5x+1}{x^2+x-2}$. The first step is to factorise the denominator: $x^2+x-2 = (x+2)(x-1)$, so we can write
| $\dfrac{5x+1}{(x+2)(x-1)}$ | $=$ | $\dfrac{A}{x+2}+\dfrac{B}{x-1} = \dfrac{A(x-1)+B(x+2)}{(x+2)(x-1)}$ |
Cancelling the denominators gives $5x+1 = A(x-1)+B(x+2)$ (1). Now let $x=-2$ so the $B$ term vanishes: $-9 = -3A$, so $A=3$. Let $x=1$ so the $A$ term vanishes: $6 = 3B$, so $B=2$.
Sanity check: substitute into (1): $3(x-1)+2(x+2) = 5x+1$ ✓. So $\dfrac{5x+1}{x^2+x-2} = \dfrac{3}{x+2}+\dfrac{2}{x-1}$.
Example 1: Simplify $\dfrac{17x+8}{6x^2+7x-20}$.
Factorise: $6x^2+7x-20 = (2x+5)(3x-4)$, so $\dfrac{17x+8}{6x^2+7x-20} = \dfrac{A}{2x+5}+\dfrac{B}{3x-4}$ and $17x+8 = A(3x-4)+B(2x+5)$.
Letting $x=-\tfrac{5}{2}$ gives $A=3$; letting $x=\tfrac{4}{3}$ gives $B=4$.
Sanity check: $3(3x-4)+4(2x+5) = 9x-12+8x+20 = 17x+8$ ✓
Repeated Factors
If a denominator has a repeated factor we need a partial fraction for each power of the factor. An expression of the form $\dfrac{x+a}{(x+b)^2}$ decomposes as $\dfrac{A}{x+b}+\dfrac{B}{(x+b)^2}$.
Example 2: Simplify $\dfrac{-21x-12}{49x^2+28x+4}$.
The denominator factorises as $(7x+2)^2$, so $\dfrac{-21x-12}{(7x+2)^2} = \dfrac{A}{7x+2}+\dfrac{B}{(7x+2)^2}$ and $-21x-12 = A(7x+2)+B$.
Comparing coefficients: $-21 = 7A$ so $A=-3$; and $-12 = 2A+B$ so $B=-6$.
Sanity check: $-3(7x+2)-6 = -21x-12$ ✓. So the answer is $\dfrac{-3}{7x+2}-\dfrac{6}{(7x+2)^2}$.
Non-linear Factors
If a denominator contains an irreducible quadratic factor $ax^2+bx+c$ (no real roots), its numerator must include terms in $x$ up to one order less than the factor — that is, $\dfrac{Ax+B}{ax^2+bx+c}$.
Example 3: Simplify $\dfrac{-116x+112}{(x^2+3x-3)(8x+4)}$.
The quadratic has no real roots, so write $\dfrac{-116x+112}{(x^2+3x-3)(8x+4)} = \dfrac{Ax+B}{x^2+3x-3}+\dfrac{C}{8x+4}$, giving
$-116x+112 = (Ax+B)(8x+4)+C(x^2+3x-3)$ (1).
Comparing coefficients: $0 = 8A+C$, $-116 = 4A+8B+3C$, $112 = 4B-3C$. Solving gives $C=-40$, $A=5$, $B=-2$.
Sanity check: $(5x-2)(8x+4)-40(x^2+3x-3) = -116x+112$ ✓. So the answer is $\dfrac{5x-2}{x^2+3x-3}-\dfrac{40}{8x+4}$.
Proper and Improper Fractions
When the numerator's order is the same as (or higher than) the denominator's, first divide out a polynomial part, then find partial fractions for the proper remainder. For example, $\dfrac{x^2-3x+2}{x^2+2x+1}$ can be written $1+\dfrac{-5x+1}{x^2+2x+1}$, and the fractional part is then handled as above.
Example 4: Simplify $\dfrac{16x^2-34x-29}{8x^2-2x-15}$.
Since both are degree 2, divide out: $\dfrac{16x^2-34x-29}{8x^2-2x-15} = \dfrac{2(8x^2-2x-15)-30x+1}{8x^2-2x-15} = 2-\dfrac{30x-1}{8x^2-2x-15}$.
Factorise $8x^2-2x-15 = (2x-3)(4x+5)$, so $30x-1 = A(4x+5)+B(2x-3)$, giving $A=4$ and $B=7$.
Sanity check: $4(4x+5)+7(2x-3) = 16x+20+14x-21 = 30x-1$ ✓. So the answer is $2-\dfrac{4}{2x-3}-\dfrac{7}{4x+5}$.