Partial Fractions (DP IB Analysis & Approaches (AA): HL): Revision Note

Partial fractions

What are partial fractions?

• Partial fractions are the reverse of adding or subtracting algebraic fractions

  • e.g. $\frac{5x}{\left(x-2\right)\left(x+1\right)}$ in partial fractions is $\frac{2}{x-2}+\frac{3}{x+1}$

    • This reverses $\frac{2}{x-2}+\frac{3}{x+1}=\frac{5x}{\left(x-2\right)\left(x+1\right)}$

How do I find partial fractions when the denominator is quadratic?

• STEP 1
  Factorise the denominator into two linear factors (and simplify if necessary)

  • e.g.  $\frac{5x+5}{x^2 +x-6}=\frac{5x+5}{(x+3)(x-2)}$

• STEP 2
  Split the fraction into the sum of two smaller fractions with linear denominators and unknown constant numerators

  • Use $A$ and $B$ to represent the unknown numerators

    • e.g.  $\frac{5x+5}{(x+3)(x-2)}\equiv \frac{A}{x+3}+\frac{B}{x-2}$

• STEP 3
  Multiply both sides by the denominator to eliminate fractions

  • e.g. $\cancel{(x+3)}\cancel{(x-2)}\times \frac{5x+5}{\cancel{(x+3)}\cancel{(x-2)}}=\cancel{(x+3)}(x-2)\times \frac{A}{\cancel{x+3}}+(x+3)\cancel{(x-2)}\times \frac{B}{\cancel{x-2}}$

  • which gives $5x+5=A\left(x-2\right)+B(x+3)$

• STEP 4
  Find the unknown constants by substituting different numerical values into both sides to form and solve simultaneous equations in $A$ and $B$

  • The easiest equations come from substituting in the roots of each linear factor

    • e.g. let $x=2$:  $5\left(2\right)+5 =A\left(2-2\right)+B\left(2+3\right)=5B$

    • so $15=5B$ giving $B=3$

    • Let $x=-3$: $5\left(-3\right)+5=A\left(-3-2\right)+B\left(-3+3\right)=-5A$

    • so $-10=-5A$ giving $A=2$

  • An alternative method is comparing coefficients

    • e.g. write it as  $5x+5 =\left(A+B\right)x+(-2A+3B)$

    • The coefficient of $x$ on both sides must match, so $5=A+B$

    • The constant term on both sides must match, so $5=-2A+3B$

    • Then solve these simultaneously

• STEP 5
  Write the original fraction in partial fractions

  • Substitute $A$ and $B$ back into the expression

    • e.g.  $\frac{5x+5}{x^2+x-6}=\frac{2}{x+3}+\frac{3}{x-2}$

• In general, if the denominator factorises then

  • $\frac{ax+b}{(cx+d)(ex+f)}=\frac{A}{cx+d}+\frac{B}{ex+f}$

How do I find partial fractions when the denominator is the square of a linear term?

• A squared linear factor in the denominator must be split into two different partial fractions of the form $\frac{A}{ax+b}+ \frac{B}{{\left(ax+b\right)}^2}$

  • This can be seen in reverse by trying to add, for example, $\frac{1}{x-3}+\frac{4}{{\left(x-3\right)}^2}$

    • The lowest common denominator is ${\left(x-3\right)}^2$

    • $\frac{1}{x-3}+\frac{4}{{\left(x-3\right)}^2}=...=\frac{x+1}{{\left(x-3\right)}^2}$

• In general, $\frac{c}{{\left(ax+b\right)}^2}=\frac{A}{ax+b}+\frac{B}{{\left(ax+b\right)}^2}$

  • Then use the same method as above

How do I find partial fractions when both the numerator and denominator are linear?

• If the numerator and denominator are both linear, the fraction can be split into a constant and a simpler fraction

  • $\frac{ax+b}{cx+d}=A+\frac{B}{cx+d}$

    • Then use the same method as above

• e.g. write$\frac{12x-2}{3x-1}$ as $A+\frac{B}{3x-1}$ to get $12x-2 =A\left(3x-1\right)+B$

  • Substitute in $x=\frac{1}{3}$ to get $B=2$

  • Substitute in, say, $x=0$ to get $-2=-A+2$ so $A=4$

    • $\frac{12x-2}{3x-1}=4+\frac{2}{3x-1}$