Simple 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. 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 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. | ||
Repeated Factors
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 $\dfrac{A}{(x+b)}+\dfrac{B}{(x+b)^2}$.
Example 2: Simplify the following expression $\dfrac{-21x-12}{49x^2+28x+4}$
| You can probably see $49x^2+28x+4$ can be factorised as $(7x+2)^2$ so we can write: | ||
| $\dfrac{-21x-12}{49x^2+28x+4}$ | $=$ | $\dfrac{A}{7x+2}+\dfrac{B}{(7x+2)^2}$ |
| So we can write: | ||
| $-21x-12$ | $=$ | $A(7x+2)+B$ (1) |
| Comparing coefficients we get: | ||
| $-21$ | $=$ | $7A$ |
| so $A$ | $=$ | $-3$ |
| $-12$ | $=$ | $2A+B$ |
| so $B$ | $=$ | $-6$ |
| Sanity Check: Insert these values for $A$ and $B$ into equation (1) | ||
| $-21x-12$ | $=$ | $-3(7x+2)-6$ |
| $=$ | $-21x-12$ | |
| so we can be confident our values for $A$ and $B$ are correct. | ||
| Finally we can write: | ||
| $\dfrac{-21x-12}{49x^2+28x+4}$ | $=$ | $\dfrac{-3}{7x+2}-\dfrac{6}{(7x+2)^2}$  |
Non-linear Factors
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 $\dfrac{Ax+B}{ax^2+bx+c}$.
Example 3: Simplify the following expression $\dfrac{-116x+112}{(x^2+3x-3)(8x+4)}$
| If you look at the quadratic term in the denominator you can see there are no real roots which means we have to treat it as a non-linear factor. | ||
| Write the expresion like this: | ||
| $\dfrac{-116x+112}{(x^2+3x-3)(8x+4)}$ | $=$ | $\dfrac{Ax+B}{x^2+3x-3}+\dfrac{C}{8x+4}$ |
| Putting the right hand side over a common denominator and then cancelling the denominators we get: | ||
| $-116x+112$ | $=$ | $(Ax+B)(8x+4)+C(x^2+3x-3)$ |
| Call this equation (1) | ||
| Comparing the coefficients of $x$ we get: | ||
| $0$ | $=$ | $8A+C$ (2) |
| $-116$ | $=$ | $4A+8B+3C$ (3) |
| $112$ | $=$ | $4B-3C$ (4) |
| From (2) $A$ | $=$ | $-C/8$ |
| From (4) $B$ | $=$ | $(112+3C)/4$ |
| Substituting in to (3) we get: | ||
| $-116$ | $=$ | $-C/2+2(112+3C)+3C$ |
| $-232$ | $=$ | $-C+4(112+3C)+6C$ |
| $-680$ | $=$ | $17C$ |
| so $C$ | $=$ | $-40$ |
| and $A$ | $=$ | $5$ |
| $B$ | $=$ | $-2$ |
| Sanity Check: Insert these values for $A$, $B$ and $C$ into equation (1) | ||
| $-116x+112$ | $=$ | $(5x-2)(8x+4)-40(x^2+3x-3)$ |
| $=$ | $40x^2+20x-16x-8$ | |
| $-40x^2-120x+120$ | ||
| $=$ | $-116x+112$ | |
| which means we can be confident our values for $A$, $B$ and $C$ are correct. | ||
| So our answer is: | ||
| $\dfrac{-116x+112}{(x^2+3x-3)(8x+4)}$ | $=$ | $\dfrac{5x-2}{x^2+3x-3}-\dfrac{40}{8x+4}$ |
Proper and Improper Fractions
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 of factorisation to derive a quotient where the numerator is at leats one order less than the denominator. In general terms this will give a non-fractional and a fractional expression of $x$. You can then find partial fractions for the fractional expression.
Example 4: Simplify the following expression $\dfrac{16x^2-34x-29}{8x^2-2x-15}$
| There is $16x^2$ in the numerator and $8x^2$ in the denominator so lets rewrite the expression like this: | ||
| $\dfrac{16x^2-34x-29}{8x^2-2x-15}$ | $=$ | $\dfrac{2(8x^2-2x-15)-30x+1}{8x^2-2x-15}$ |
| $=$ | $2-\dfrac{30x-1}{8x^2-2x-15}$ | |
| You can check the expression is correct by collecting common terms in the numerator. Be careful with the minus signs. | ||
| Next we decompose $\dfrac{30x-1}{8x^2-2x-15}$ in the normal way. Factorise the denominator. | ||
| ${8x^2-2x-15}$ | $=$ | $(2x-3)(4x+5)$ |
| So we can write: | ||
| $30x-1$ | $=$ | $A(4x+5)+B(2x-3)$ (1) |
| Solving for $A$ and $B$ gives $A=4$ and $B=7$ | ||
| Sanity Check: Insert these values for $A$ and $B$ into equation (1) | ||
| $4(4x+5)+7(2x-3)$ | $=$ | $16x+20+14x-21$ |
| $=$ | $30x-1$ | |
| so we can be confident our values for $A$ and $B$ are correct. | ||
| So finally we can write: | ||
| $\dfrac{16x^2-34x-29}{8x^2-2x-15}$ | $=$ | $2-\dfrac{4}{2x-3}-\dfrac{7}{4x+5}$ |