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. You can probably see that | ||

$x^2+x-2$ | $=$ | $(x+2)(x-1)$ |

so we can write | ||

$\dfrac{5x+1}{x^2+x-2}$ | $=$ | $\dfrac{5x+1}{(x+2)(x-1)}$ |

Imagine there are two values, $A$ and $B$ such that | ||

$\dfrac{5x+1}{x^2+x-2}$ | $=$ | $\dfrac{A}{x+2}+\dfrac{B}{x-1}$ |

Putting the right hand side over a common denominator | ||

$\dfrac{5x+1}{x^2+x-2}$ | $=$ | $\dfrac{A(x-1)+B(x+2)}{(x+2)(x-1)}$ |

Cancelling the denominators | ||

$5x+1$ | $=$ | $A(x-1)+B(x+2)$ (1) |

We have called this equation (1) because we will use it later for a sanity check. Now, let $x=-2$ so the $B$ term becomes zero. | ||

$5(-2)+1$ | $=$ | $A(-2-1)+B(-2+2)$ |

$-9$ | $=$ | $-3A$ |

$A$ | $=$ | $3$ |

Next, let $x=1$ so the $A$ term becomes zero. | ||

$5 \times 1+1$ | $=$ | $A(1-1)+B(1+2)$ |

$B$ | $=$ | $2$ |

For a sanity check we will substitute these values of $A$ and $B$ into equation (1). | ||

$3(x-1)+2(x+2)$ | $=$ | $5x+1$ |

This agrees with equation (1) so we can be confident $A$ and $B$ are correct. |

**Example 6.1**: Simplify the following expression $\dfrac{17x+8}{6x^2+7x-20}$

Start the process of simplification by factorising the denominator. | ||

$6x^2+7x-20$ | $=$ | $(2x+5)(3x-4)$ |

So we can write: | ||

$\dfrac{17x+8}{6x^2+7x-20}$ | $=$ | $\dfrac{A}{2x+5}+\dfrac{B}{3x-4}$ |

Put the right hand side over a common denominator | ||

$\dfrac{17x+8}{6x^2+7x-20}$ | $=$ | $\dfrac{A(3x-4)+B(2x+5)}{(2x+5)(3x-4)}$ |

so $17x+8$ | $=$ | $A(3x-4)+B(2x+5)$ (1) |

If we let $x=-5/2$ we can evaluate $A=3$. If we let $x=4/3$ $B=4$. | ||

Sanity Check: Insert these values for $A$ and $B$ into equation (1) | ||

$A(3x-4)+B(2x+5)$ | $=$ | $3(3x-4)+4(2x+5)$ |

$=$ | $9x-12+8x+20$ | |

$=$ | $17x+8$ | |

so we can be confident our values for $A$ and $B$ are correct. |

If a fraction has repeated factors in the denominator we need partial fractions that account for all the possible factors. To do this we put $A$ over the factor, $B$ over the square of the factor and so on until we have all the possible factors. For example, an expression of the form $\dfrac{x+a}{(x+b)^2}$ would be factorised like this $\dfrac{A}{(x+b)}+\dfrac{B}{(x+b)^2}$.

$\dfrac{x+a}{(x+b)^2}=\dfrac{A}{(x+b)}+\dfrac{B}{(x+b)^2}$.

If a fraction contains a non-linear factor of the form $ax^2+bx+c$ we need a numerator that includes terms in $x$ up to an order one less than the factor in the denominator. The partial fraction for a factor of the form $ax^2+bx+c$ would be $\dfrac{Ax+B}{ax^2+bx+c}$.

$\dfrac{x+a}{ax^2+bx+c}=\dfrac{Ax+B}{ax^2+bx+c}$.

In cases where the order of the numerator is the same as the order of the denominator extract a constant term and then find partial fractions for the remainder. For example if we have the fraction $\dfrac{x^2-3x+2}{x^2+2x+1}$ the numerator can be rewritten as $x^2+2x+1+(-5x+1)$ which means the fraction can be written as $\dfrac{x^2-3x+2}{x^2+2x+1}=1+\dfrac{(-5x+1)}{x^2+2x+1}$. The fractional part can now be solved as above.

In cases where the order of the numerator is higher than the order of the denominator we use algebraic division to divide the numerator by the denominator. In general terms this will give a non-fractional function of $x$ and a fractional remainder. You can then find partial fractions for the remainder.