Partial fractions

​​In a nutshell

An algebraic fraction can be written in the form $$\dfrac{P(x)}{Q(x)}$$, where $$P(x)$$ and $$Q(x)$$ are polynomials. The order of a polynomial is given by the highest power of $$x$$. If the order of $$P(x)$$ is less than the order of $$Q(x)$$, then it is a proper fraction, and can be split up into two or more partial fractions. The denominators of the partial fractions are based on the linear factors of $$Q(x)$$.​

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Partial fractions

When splitting a algebraic fraction into partial fractions, there are two methods, which are:

• Substitution
  
• Equating coefficients

Substitution

procedure

Example 1

Split $$\cfrac{7x+4}{x^2-49}$$ into partial fractions using the substitution method.

First, factorise the denominator into linear factors.

$$\cfrac{7x+4}{x^2-49}=\cfrac{7x+4}{(x+7)(x-7)}$$​​

Now, set $$\it A$$ and $$\it B$$ as numerators of the partial fractions.

​$$\cfrac{7x+4}{(x+7)(x-7)} = \cfrac{A}{x+7}+\cfrac{B}{x-7}$$​​

Multiply up the fractions to eliminate the denominators.

​$$\begin{aligned}\dfrac{7x+4}{(x+7)(x-7)} &= \dfrac{A}{x+7}+\dfrac{B}{x-7}\\\dfrac{7x+4}{(x+7)(x-7)} &=\cfrac{A(x-7)}{(x+7)(x-7)} + \cfrac{B(x+7)}{(x+7)(x-7)} \\\\7x+4 & = A(x-7) + B(x+7)\end {aligned}$$​​

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In order to find $$A$$, substitute $$x=-7$$, so $$B$$ is eliminated from the equation.

​$$\begin{aligned}{7(-7)+4} &={A(-7-7) + B(-7+7)} \\-49 +4 &= -14A \\ A&= \underline{\dfrac {45}{14}}\end{aligned}$$​​

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Repeat the process using $$x=7$$​ in order to find $$B$$.

​$$\begin{aligned}7 \times 7 + 4 &= A ( 7-7) + B (7+7) \\ 53 &= 14B \\ B&= \underline{ \dfrac{53}{14}}\end{aligned}$$​​

​$$53= 14B$$

$$\underline{B = \cfrac{53}{14}}$$​​​

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Therefore, $$\underline{\dfrac{7x+4}{x^2-49}=\dfrac{45}{14(x+7)} + \dfrac{53}{14(x-7)}}$$​​

Equating coefficients

procedure

Example 2

Split $$\cfrac{7x+4}{x^2-49}$$ into partial fractions using the equating coefficients method. 

First, factorise the denominator into linear factors.

$$\cfrac{7x+4}{x^2-49} = \cfrac{7x+4}{(x+7)(x-7)}$$​​

Now, set $$\it A$$ and $$\it B$$ as numerators of the partial fractons.

$$\cfrac{7x+4}{(x+7)(x-7)}=\cfrac{A}{x+7}+\cfrac{B}{x-7}$$

Multiply up the fractions to eliminate the denominators.

$$\begin{aligned}\dfrac{7x+4}{(x+7)(x-7)} &= \dfrac{A}{x+7}+\dfrac{B}{x-7}\\\dfrac{7x+4}{(x+7)(x-7)} &=\cfrac{A(x-7)}{(x+7)(x-7)} + \cfrac{B(x+7)}{(x+7)(x-7)} \\\\7x+4 & = A(x-7) + B(x+7)\end {aligned}$$​​​

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Next, expand all the brackets, to find the coefficients.

$$7x+4 = Ax - 7A + Bx + 7B$$​​

$$7x+4 = x(A+B)+7(B-A)$$​​

Equate coefficients of x.

$$7x = x(A+B)$$​​

And constant terms.

$$4 = 7 (B-A)$$​​

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And solve both equations simultaneously.

​$$\begin{cases}7x = x(A+B) \\4= 7 (B-A) \end{cases}$$​​

Giving:

​$$\underline{A=\cfrac{45}{14}} \hspace{10mm}\underline{B=\cfrac{53}{14}}$$

Note: You will have to split your fractions into as many terms ($$A, \ B, \ C$$) as factors in the denominator you have.