What you'll learn
- How to recognise a rational function and decide whether algebraic division is needed.
- How to choose the correct partial-fraction form for distinct, repeated and quadratic factors.
- How to find unknown coefficients by substitution or by comparing coefficients.
- How to check that a decomposition is correct.
Prerequisites
You should be comfortable expanding brackets, factorising polynomials, solving simultaneous equations and dividing polynomials.
A polynomial is an expression involving non-negative integer powers of a variable, such as 2x3−x+52x^3-x+52x3−x+5.
Rational function
A rational function is a fraction of the form
P(x)Q(x),\frac{P(x)}{Q(x)},Q(x)P(x),where P(x)P(x)P(x) and Q(x)Q(x)Q(x) are polynomials and Q(x)≠0Q(x)\neq 0Q(x)=0.
What are partial fractions?
A complicated rational function can sometimes be rewritten as a sum of simpler fractions. These simpler fractions are called its partial fractions.
For example, an expression with denominator (x+1)(x+3)(x+1)(x+3)(x+1)(x+3) may be written in the form
Ax+1+Bx+3,\frac{A}{x+1}+\frac{B}{x+3},x+1A+x+3B,where AAA and BBB are constants to be found.
Decomposition
To decompose a rational function means to rewrite it as a sum of simpler rational functions. The denominator's factors determine the required form.
Partial fractions are particularly useful when integrating rational functions, although you can also be asked simply to find the decomposition.
Proper and improper rational functions
Proper rational function
A rational function is proper if the degree of its numerator is less than the degree of its denominator. The degree is the highest power of xxx present.
For example, 3x+1x2+4x+3\frac{3x+1}{x^2+4x+3}x2+4x+33x+1 is proper because the numerator has degree 1 and the denominator has degree 2.
A rational function is improper if the numerator's degree is greater than or equal to the denominator's degree. You must use polynomial division before decomposing an improper rational function.
Making an improper fraction proper
Decompose
x2+4x+1x2+x−2.\frac{x^2+4x+1}{x^2+x-2}.x2+x−2x2+4x+1.-
The numerator and denominator both have degree 2, so begin with polynomial division. Since their leading terms are both x2x^2x2, the quotient starts with 1:
x2+4x+1x2+x−2=1+3x+3x2+x−2.\frac{x^2+4x+1}{x^2+x-2} =1+\frac{3x+3}{x^2+x-2}.x2+x−2x2+4x+1=1+x2+x−23x+3. -
Factorise the denominator of the proper remainder:
x2+x−2=(x−1)(x+2).x^2+x-2=(x-1)(x+2).x2+x−2=(x−1)(x+2).Therefore write
3x+3(x−1)(x+2)=Ax−1+Bx+2.\frac{3x+3}{(x-1)(x+2)} =\frac{A}{x-1}+\frac{B}{x+2}.(x−1)(x+2)3x+3=x−1A+x+2B. -
Multiply through by (x−1)(x+2)(x-1)(x+2)(x−1)(x+2):
3x+3=A(x+2)+B(x−1).3x+3=A(x+2)+B(x-1).3x+3=A(x+2)+B(x−1).Substituting x=1x=1x=1 gives 6=3A6=3A6=3A, so A=2A=2A=2. Substituting x=−2x=-2x=−2 gives −3=−3B-3=-3B−3=−3B, so B=1B=1B=1.
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Hence
x2+4x+1x2+x−2=1+2x−1+1x+2.\frac{x^2+4x+1}{x^2+x-2} =1+\frac{2}{x-1}+\frac{1}{x+2}.x2+x−2x2+4x+1=1+x−12+x+21.
Skipping algebraic division
Do not begin a partial-fraction decomposition while the rational part is improper. Divide first, even when the numerator and denominator have the same degree.
Distinct linear factors
A linear factor has the form ax+bax+bax+b. Distinct linear factors are different factors, such as x−2x-2x−2 and x+3x+3x+3.
If the denominator contains two distinct linear factors, use
P(x)(ax+b)(cx+d)=Aax+b+Bcx+d.\frac{P(x)}{(ax+b)(cx+d)} =\frac{A}{ax+b}+\frac{B}{cx+d}.(ax+b)(cx+d)P(x)=ax+bA+cx+dB.The numerator above each linear factor is a constant because its degree must be less than the degree of that factor.
Decomposing over two distinct linear factors
Decompose
5x+1(x−1)(x+2).\frac{5x+1}{(x-1)(x+2)}.(x−1)(x+2)5x+1.-
Choose the form
5x+1(x−1)(x+2)=Ax−1+Bx+2.\frac{5x+1}{(x-1)(x+2)} =\frac{A}{x-1}+\frac{B}{x+2}.(x−1)(x+2)5x+1=x−1A+x+2B.Multiplying through by the full denominator gives
5x+1=A(x+2)+B(x−1).5x+1=A(x+2)+B(x-1).5x+1=A(x+2)+B(x−1). -
Substitute x=1x=1x=1 to eliminate the term containing BBB:
6=3A,6=3A,6=3A,so A=2A=2A=2.
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Substitute x=−2x=-2x=−2 to eliminate the term containing AAA:
−9=−3B,-9=-3B,−9=−3B,so B=3B=3B=3.
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Therefore
5x+1(x−1)(x+2)=2x−1+3x+2.\frac{5x+1}{(x-1)(x+2)} =\frac{2}{x-1}+\frac{3}{x+2}.(x−1)(x+2)5x+1=x−12+x+23.
Choose convenient substitutions
After clearing the denominator, substitute the roots of its linear factors. Each root usually makes all but one unknown term equal to zero.
Three distinct linear factors
If the denominator contains three distinct linear factors, include one fraction for each factor:
P(x)(x−a)(x−b)(x−c)=Ax−a+Bx−b+Cx−c.\frac{P(x)}{(x-a)(x-b)(x-c)} =\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}.(x−a)(x−b)(x−c)P(x)=x−aA+x−bB+x−cC.This is the largest number of partial-fraction terms required in this specification.
Decomposing into three terms
Decompose
2x2+5x−1(x−1)(x+1)(x+2).\frac{2x^2+5x-1}{(x-1)(x+1)(x+2)}.(x−1)(x+1)(x+2)2x2+5x−1.-
Write
2x2+5x−1(x−1)(x+1)(x+2)=Ax−1+Bx+1+Cx+2.\frac{2x^2+5x-1}{(x-1)(x+1)(x+2)} =\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x+2}.(x−1)(x+1)(x+2)2x2+5x−1=x−1A+x+1B+x+2C.Clearing the denominator gives
2x2+5x−1=A(x+1)(x+2)+B(x−1)(x+2)+C(x−1)(x+1).2x^2+5x-1 =A(x+1)(x+2)+B(x-1)(x+2)+C(x-1)(x+1).2x2+5x−1=A(x+1)(x+2)+B(x−1)(x+2)+C(x−1)(x+1). -
Substitute x=1x=1x=1:
6=6A,6=6A,6=6A,so A=1A=1A=1.
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Substitute x=−1x=-1x=−1:
−4=−2B,-4=-2B,−4=−2B,so B=2B=2B=2.
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Substitute x=−2x=-2x=−2:
−3=3C,-3=3C,−3=3C,so C=−1C=-1C=−1. Hence
2x2+5x−1(x−1)(x+1)(x+2)=1x−1+2x+1−1x+2.\frac{2x^2+5x-1}{(x-1)(x+1)(x+2)} =\frac{1}{x-1}+\frac{2}{x+1}-\frac{1}{x+2}.(x−1)(x+1)(x+2)2x2+5x−1=x−11+x+12−x+21.
Repeated linear factors
A repeated factor is a factor that occurs more than once. If the denominator contains (x−a)2(x-a)^2(x−a)2, you must include terms for both powers:
P(x)(x−a)2(x−b)=Ax−a+B(x−a)2+Cx−b.\frac{P(x)}{(x-a)^2(x-b)} =\frac{A}{x-a}+\frac{B}{(x-a)^2}+\frac{C}{x-b}.(x−a)2(x−b)P(x)=x−aA+(x−a)2B+x−bC.Every power is needed
For a repeated factor (ax+b)2(ax+b)^2(ax+b)2, include one fraction over ax+bax+bax+b and another over (ax+b)2(ax+b)^2(ax+b)2.
Decomposing with a repeated factor
Decompose
4x2−x−1(x−1)2(x+1).\frac{4x^2-x-1}{(x-1)^2(x+1)}.(x−1)2(x+1)4x2−x−1.-
Use the full repeated-factor form:
4x2−x−1(x−1)2(x+1)=Ax−1+B(x−1)2+Cx+1.\frac{4x^2-x-1}{(x-1)^2(x+1)} =\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+1}.(x−1)2(x+1)4x2−x−1=x−1A+(x−1)2B+x+1C.Clearing the denominator gives
4x2−x−1=A(x−1)(x+1)+B(x+1)+C(x−1)2.4x^2-x-1=A(x-1)(x+1)+B(x+1)+C(x-1)^2.4x2−x−1=A(x−1)(x+1)+B(x+1)+C(x−1)2. -
Substitute x=1x=1x=1:
2=2B,2=2B,2=2B,so B=1B=1B=1. Substitute x=−1x=-1x=−1:
4=4C,4=4C,4=4C,so C=1C=1C=1.
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To find AAA, choose an easy value such as x=0x=0x=0:
−1=−A+1+1,-1=-A+1+1,−1=−A+1+1,giving A=3A=3A=3.
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Therefore
4x2−x−1(x−1)2(x+1)=3x−1+1(x−1)2+1x+1.\frac{4x^2-x-1}{(x-1)^2(x+1)} =\frac{3}{x-1}+\frac{1}{(x-1)^2}+\frac{1}{x+1}.(x−1)2(x+1)4x2−x−1=x−13+(x−1)21+x+11.
Omitting the lower power
Writing only A(x−a)2\frac{A}{(x-a)^2}(x−a)2A is incomplete. A squared linear factor requires terms with denominators x−ax-ax−a and (x−a)2(x-a)^2(x−a)2.
Irreducible quadratic factors
An irreducible quadratic factor is a quadratic that cannot be factorised into real linear factors. For example, x2+4x^2+4x2+4 is irreducible over the real numbers.
Above a quadratic factor, the numerator must be linear:
P(x)(x−a)(x2+b)=Ax−a+Bx+Cx2+b.\frac{P(x)}{(x-a)(x^2+b)} =\frac{A}{x-a}+\frac{Bx+C}{x^2+b}.(x−a)(x2+b)P(x)=x−aA+x2+bBx+C.The numerator Bx+CBx+CBx+C has lower degree than the quadratic denominator but is general enough to represent all possibilities.
Decomposing with a quadratic factor
Decompose
2x2+3x+5(x−1)(x2+4).\frac{2x^2+3x+5}{(x-1)(x^2+4)}.(x−1)(x2+4)2x2+3x+5.-
Write
2x2+3x+5(x−1)(x2+4)=Ax−1+Bx+Cx2+4.\frac{2x^2+3x+5}{(x-1)(x^2+4)} =\frac{A}{x-1}+\frac{Bx+C}{x^2+4}.(x−1)(x2+4)2x2+3x+5=x−1A+x2+4Bx+C.Clearing the denominator gives
2x2+3x+5=A(x2+4)+(Bx+C)(x−1).2x^2+3x+5=A(x^2+4)+(Bx+C)(x-1).2x2+3x+5=A(x2+4)+(Bx+C)(x−1). -
Substitute x=1x=1x=1:
10=5A,10=5A,10=5A,so A=2A=2A=2.
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Expand the identity using A=2A=2A=2:
2x2+3x+5=2x2+8+Bx2+(C−B)x−C.2x^2+3x+5 =2x^2+8+Bx^2+(C-B)x-C.2x2+3x+5=2x2+8+Bx2+(C−B)x−C.Comparing coefficients of x2x^2x2 gives B=0B=0B=0, and comparing coefficients of xxx gives C−B=3C-B=3C−B=3, so C=3C=3C=3.
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Therefore
2x2+3x+5(x−1)(x2+4)=2x−1+3x2+4.\frac{2x^2+3x+5}{(x-1)(x^2+4)} =\frac{2}{x-1}+\frac{3}{x^2+4}.(x−1)(x2+4)2x2+3x+5=x−12+x2+43.
Comparing coefficients and checking
An equation that is true for every permitted value of xxx is called an identity. Once denominators have been cleared, you may expand both sides and compare the coefficients of matching powers of xxx.
A quick check is to recombine your final fractions over their common denominator. Their numerator should simplify to the original numerator.
Use a combined method
Use factor roots to find coefficients quickly, then substitute another convenient value or compare coefficients to find any remaining unknowns.
In the exam
- Check whether the rational function is proper; if it is not, use polynomial division first.
- Factorise the denominator completely, then write every required partial-fraction term before finding any coefficients.
- Clear the denominator carefully and use roots of linear factors as substitutions.
- For repeated factors or quadratic factors, use another convenient value or compare coefficients.
- Check the result by recombining the fractions or testing a permissible value of xxx.
Check yourself
- What partial-fraction form is required when the denominator is (x−2)2(x+5)(x-2)^2(x+5)(x−2)2(x+5)?
- Why must the numerator above an irreducible quadratic factor be written as Ax+BAx+BAx+B rather than just a constant?
- What must you do before decomposing a rational function whose numerator and denominator have the same degree?
