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Reference Sheet: Programme F.8 — Partial Fractions

Rules, procedures and drill, in the order of the book's programme. Every rule links to the unit in the lesson that derives it — if a rule here looks arbitrary, the link is where it stops being arbitrary.

The book's vocabulary is used throughout: algebraic fraction, prime factors, simple factor (ax+b), irreducible quadratic factor, equating coefficients, \equiv for an identity. Our own terms are given in brackets, and material that is not in the book is marked Beyond the book.

1. Partial fractions: adding fractions run backwards

From Unit 01.

Fractions are added over a common denominator, the LCM of their denominators. The simple fractions that were added are the partial fractions of the result; finding them is the addition run backwards.

Example. \frac2{x-3}-\frac4{x-1}=\frac{2(x-1)-4(x-3)}{(x-3)(x-1)}=\frac{10-2x}{(x-3)(x-1)}, so \frac2{x-3} and -\frac4{x-1} are the partial fractions of \frac{10-2x}{(x-3)(x-1)}.

2. The identity: why any value of x may be substituted

From Unit 01.

Factorise the denominator into its prime factors, assume one partial fraction per factor, add them over the original denominator, and equate the numerators. The numerator equation is an identity (\equiv), true for every x, including the values the original fraction excludes: both sides are polynomials that agree at every other x, infinitely many points, and two polynomials of degree at most n that agree at n+1 points are the same polynomial.

Example. \frac{5x+1}{(x-1)(x+2)}=\frac A{x-1}+\frac B{x+2} gives 5x+1\equiv A(x+2)+B(x-1), which may be evaluated at x=1 and x=-2 even though the fraction has no value there.

3. Two simple factors: substituting suitable values

From Unit 01.

\frac{P(x)}{(x-a)(x-b)}=\frac A{x-a}+\frac B{x-b},\qquad a\ne b,\quad \deg P<2.

Substitute x=a and x=b in turn: each empties one bracket and isolates the other unknown. The split always exists and is unique. Beyond the book: this shortcut is called the cover-up rule.

Example. \frac{8x-28}{x^2-6x+8}: x^2-6x+8=(x-2)(x-4) and 8x-28\equiv A(x-4)+B(x-2). At x=4: 4=2B, so B=2. At x=2: -12=-2A, so A=6. Answer: \frac6{x-2}+\frac2{x-4}.

4. The degree condition

From Unit 02.

The book's rule: the degree of the numerator must be less than the degree of the denominator (in our words, the fraction must be proper). If it is not (an improper fraction), divide out by long division first (F.2 Unit 05).

5. Improper fractions: a constant quotient

From Unit 02.

\frac{P(x)}{Q(x)}=S(x)+\frac{R(x)}{Q(x)},\qquad \deg R<\deg Q,

then split \frac{R(x)}{Q(x)} as usual and keep the quotient S(x). When the numerator and denominator have the same degree, the quotient is a number.

Examples. The book's Example 1: \frac{x^2+3x-10}{x^2-2x-3}=1+\frac{5x-7}{(x+1)(x-3)}=1+\frac3{x+1}+\frac2{x-3}. The book's Example 2: 2x^2+18x+31=2(x^2+5x+6)+(8x+19), so \frac{2x^2+18x+31}{x^2+5x+6}=2+\frac{8x+19}{(x+2)(x+3)}. Then 8x+19\equiv A(x+3)+B(x+2); at x=-2, 3=A; at x=-3, -5=-B, so B=5. Answer: 2+\frac3{x+2}+\frac5{x+3}.

6. Improper fractions: a polynomial quotient

From Unit 02.

Whenever \deg P\ge\deg Q the fraction must be divided first, and the quotient has degree \deg P-\deg Q: a number when the degrees are equal (§5), a linear polynomial when the numerator's degree is one more, and so on.

Example. The book's Example 4, \frac{2x^3+3x^2-54x+50}{x^2+2x-24}, a cubic over a quadratic: long division gives quotient 2x-1 (degree 3-2=1) and remainder -4x+26. With x^2+2x-24=(x-4)(x+6), -4x+26\equiv A(x+6)+B(x-4). At x=4: 10=10A, so A=1. At x=-6: 50=-10B, so B=-5. Answer: 2x-1+\frac1{x-4}-\frac5{x+6}.

7. The procedure for simple factors ax+b

From Unit 01 and Unit 03.

The book's procedure: (a) write the denominator as a product of prime factors; (b) give each simple factor ax+b a partial fraction \frac A{ax+b} with A constant; (c) add the partial fractions over the original denominator; (d) equate the numerators; (e) substitute suitable values of x to find the constants. A factor ax+b with a\ne1 (a non-monic factor, in our words) has its root at x=-\frac ba, and that is the value to substitute.

Example. \frac{5x}{(3x+2)(x-1)}: 5x\equiv A(x-1)+B(3x+2). At x=1: 5=5B, so B=1. At x=-\frac23: -\frac{10}3=-\frac53A, so A=2. Answer: \frac2{3x+2}+\frac1{x-1}.

8. Irreducible quadratic factors: the test

From Unit 05.

A quadratic factor ax^2+bx+c with whole-number coefficients is irreducible (the book's term) when b^2-4ac is not a perfect square: then its roots \frac{-b\pm\sqrt{b^2-4ac}}{2a} are irrational or complex, and it has no simple factors with rational coefficients. The test is not "b^2-4ac<0"; that is only the special case with no real roots.

Examples. 3x^2+4x-2: b^2-4ac=16+24=40. 2x^2-6x+3: b^2-4ac=36-24=12. Neither is a perfect square, so both are irreducible, although both have real (irrational) roots. x^2+1: b^2-4ac=-4, irreducible with no real roots. By contrast x^2-5x+6: 25-24=1=1^2, so it splits, as (x-2)(x-3).

9. Irreducible quadratic factors: the template, by equating coefficients

From Unit 05; equating coefficients from Unit 03.

\frac{P(x)}{(px+q)(ax^2+bx+c)}=\frac A{px+q}+\frac{Bx+C}{ax^2+bx+c},\qquad \deg P<3.

A linear numerator over the quadratic, never a bare constant. To find the constants, multiply out, collect like powers, and equate coefficients, labelling the equations [x^2], [x], [\text{CT}] (constant term). Substituting x=-\frac qp still gives A directly.

Example. \frac{9x^2+48x+18}{(2x+1)(x^2+8x+3)}, with b^2-4ac=64-12=52, irreducible: 9x^2+48x+18\equiv A(x^2+8x+3)+(Bx+C)(2x+1), so [x^2]: 9=A+2B; [x]: 48=8A+B+2C; [\text{CT}]: 18=3A+C. At x=-\frac12: -\frac{15}4=-\frac34A, so A=5. Then [x^2] gives B=2, [\text{CT}] gives C=3, and [x] checks: 40+2+6=48. Answer: \frac5{2x+1}+\frac{2x+3}{x^2+8x+3}.

10. Repeated simple factors: the rule

From Unit 04.

(ax+b)^2\to\frac A{ax+b}+\frac B{(ax+b)^2},\qquad (ax+b)^3\to\frac A{ax+b}+\frac B{(ax+b)^2}+\frac C{(ax+b)^3},

one term for every power from 1 to n, none skipped (a ladder of rungs, in our words).

Examples. \frac{42x+44}{(6x+5)^2}: 42x+44\equiv A(6x+5)+B; [x]: 42=6A, so A=7; [\text{CT}]: 44=5A+B, so B=9. Answer: \frac7{6x+5}+\frac9{(6x+5)^2}. Likewise \frac{35x-14}{(7x-2)^2}=\frac5{7x-2}-\frac4{(7x-2)^2}.

11. Why every power is needed

From Unit 04.

Beyond the book (the book states the rule without a reason). With u=ax+b, a numerator of degree less than n can be rewritten as a polynomial in u of degree less than n; dividing by u^n gives exactly one term over each power u^1,\ldots,u^n. So the ladder always exists and is unique, and its numerators are the numerator's "digits" in base u. Skipping a rung forces one of those digits to be zero, which almost always contradicts the numerator. Substituting the root x=-\frac ba finds only the top rung.

Example. Skipping the middle rung of \frac{18x^2+3x+6}{(3x+1)^3} leaves A(3x+1)^2+C=9Ax^2+6Ax+(A+C): [x^2] demands A=2 and [x] demands A=0.5, so there is no solution.

12. Cubed repeated factors

From Unit 04.

Example. \frac{20x^2-54x+35}{(2x-3)^3}: 20x^2-54x+35\equiv A(2x-3)^2+B(2x-3)+C=4Ax^2+(-12A+2B)x+(9A-3B+C). [x^2]: 20=4A, so A=5. [x]: -54=-60+2B, so B=3. [\text{CT}]: 35=45-9+C, so C=-1 (and x=\frac32 gives C=45-81+35=-1 directly). Answer: \frac5{2x-3}+\frac3{(2x-3)^2}-\frac1{(2x-3)^3}. Likewise \frac{18x^2+3x+6}{(3x+1)^3}=\frac2{3x+1}-\frac3{(3x+1)^2}+\frac7{(3x+1)^3} and \frac{32x^2-28x-5}{(4x-3)^3}=\frac2{4x-3}+\frac5{(4x-3)^2}-\frac8{(4x-3)^3}.

13. Three different simple factors

From Unit 03.

One constant-numerator fraction per factor. The book equates coefficients here; substituting each root is quicker. Both give the same constants, because substitution shows they are forced (unique).

Example. \frac{10x^2+7x-42}{(x-2)(x+4)(x-1)}: substituting x=2,-4,1 gives A=2, B=3, C=5; equating coefficients gives A+B+C=10, 3A-3B+2C=7, -4A+2B-8C=-42, with the same solution. Answer: \frac2{x-2}+\frac3{x+4}+\frac5{x-1}.

14. An unfactorised cubic denominator

From Unit 05.

If the denominator arrives as a cubic, factorise it first by the remainder theorem (F.3 Unit 04): evaluate f(x) in nested form (F.3 Unit 03) at x=1,-1,2,-2,\ldots until f(k)=0, divide out (x-k) by long division, and test the quadratic left over (§8). Then split by whichever template matches the factors.

Examples. \frac{8x^2-14x-10}{x^3-4x^2+x+6}: nested f(x)=[(x-4)x+1]x+6; f(1)=4, f(-1)=0; dividing out (x+1) leaves x^2-5x+6=(x-2)(x-3), so

\frac{8x^2-14x-10}{(x+1)(x-2)(x-3)}=\frac1{x+1}+\frac2{x-2}+\frac5{x-3}.

\frac{12x^2+36x+6}{x^3+6x^2+3x-10}: nested f(x)=[(x+6)x+3]x-10 and f(1)=0; dividing out (x-1) leaves x^2+7x+10=(x+2)(x+5). Then 12x^2+36x+6\equiv A(x+2)(x+5)+B(x-1)(x+5)+C(x-1)(x+2): at x=1, 54=18A, so A=3; at x=-2, -18=-9B, so B=2; at x=-5, 126=18C, so C=7. Answer: \frac3{x-1}+\frac2{x+2}+\frac7{x+5}.

Traps

  • Forcing the template onto an improper fraction. Check the degrees first (§4). Substituting roots may even produce plausible numbers, but a sum of constants over simple factors can never supply the quotient.
  • Dropping the quotient after dividing. \frac{P(x)}{Q(x)} equals the quotient plus the split remainder, not the remainder alone.
  • Reading off A and B before clearing denominators. The unknowns stay tangled until the identity is written and a value substituted or coefficients equated.
  • Forgetting the root of a factor ax+b. It vanishes at x=-\frac ba, not at x=-b or x=b.
  • b^2-4ac<0 as the irreducibility test. The book's test is "not a perfect square" (§8); an irreducible quadratic here can have real, irrational roots.
  • A constant numerator over a quadratic or repeated factor. A quadratic factor needs Bx+C (§9), and a factor repeated n times needs n rungs (§10): one constant is too few unknowns.
  • Skipping a rung of the ladder. \frac A{ax+b}+\frac C{(ax+b)^3} alone, omitting \frac B{(ax+b)^2}, generally has no solution (§11).
  • Expecting substitution to find every constant. It finds the constant of each simple factor and the top rung of a repeated factor; the rest need equating coefficients (§9, §12).
  • Splitting a denominator that is not yet factorised. Factorise a cubic first (§14); no template applies to an unfactorised denominator.
  • Skipping the check (beyond the book: the book never recombines). Comparing both sides at one ordinary value of x, such as x=0, catches a sign or arithmetic slip that the root substitutions alone would miss.

Self-check

  1. Decompose \frac{x+7}{x^2-7x+10}.
  2. Decompose \frac{10x+37}{x^2+3x-28}.
  3. The fraction in question 1 has no value at x=2 or x=5, yet those are the values substituted into the cleared equation. Explain why that is legitimate.
  4. Decompose \frac{3x^2-8x-63}{x^2-3x-10} (check the degrees first).
  5. Decompose \frac{2x^2+6x-35}{x^2-x-12}.
  6. Decompose \frac{2x^3-x^2-3x+3}{x^2-x-2}, showing the division.
  7. Decompose \frac{x+8}{2x^2+7x+3}.
  8. Decompose \frac{5x^2+15x+1}{(x+2)(x^2+4x+1)}, after checking that the quadratic is irreducible.
  9. Show that the template \frac A{x-1}+\frac C{x^2+1} fails for \frac{x^2+2x+3}{(x-1)(x^2+1)}, then find the correct split.
  10. Decompose \frac{8x+1}{(4x+3)^2}.
  11. Decompose \frac{12x^2+8x+5}{(2x+1)^3}.
  12. Factorise x^3-2x^2-5x+6 by the remainder theorem, then decompose \frac{7x^2-6x-25}{x^3-2x^2-5x+6}.

Fully worked solutions

1. x^2-7x+10=(x-2)(x-5). Write \frac{x+7}{(x-2)(x-5)}=\frac A{x-2}+\frac B{x-5}, so x+7\equiv A(x-5)+B(x-2). At x=5: 12=3B, so B=4. At x=2: 9=-3A, so A=-3. Answer: \frac4{x-5}-\frac3{x-2}. Check at x=0: \frac7{10}=0.7 and -\frac45+\frac32=0.7.

2. x^2+3x-28=(x-4)(x+7), and 10x+37\equiv A(x+7)+B(x-4). At x=4: 77=11A, so A=7. At x=-7: -33=-11B, so B=3. Answer: \frac7{x-4}+\frac3{x+7}.

3. Clearing denominators only establishes x+7=A(x-5)+B(x-2) for x\ne2,5. Both sides are straight lines (polynomials of degree at most 1), and they agree at infinitely many points. Two lines that share two points are the same line, so the two sides agree at every x, including x=2 and x=5: the equation is an identity, x+7\equiv A(x-5)+B(x-2). With A=-3 and B=4 the right side is -3x+15+4x-8=x+7 exactly.

4. Both have degree 2, so divide: 3x^2-8x-63=3(x^2-3x-10)+(x-33), quotient 3 and remainder x-33. With x^2-3x-10=(x+2)(x-5), x-33\equiv A(x-5)+B(x+2). At x=5: -28=7B, so B=-4. At x=-2: -35=-7A, so A=5. Answer: 3+\frac5{x+2}-\frac4{x-5}.

5. Both have degree 2: 2x^2+6x-35=2(x^2-x-12)+(8x-11), quotient 2 and remainder 8x-11. With x^2-x-12=(x+3)(x-4), 8x-11\equiv A(x-4)+B(x+3). At x=4: 21=7B, so B=3. At x=-3: -35=-7A, so A=5. Answer: 2+\frac5{x+3}+\frac3{x-4}.

6. Degree 3 over degree 2, so the quotient is linear. Divide: 2x(x^2-x-2)=2x^3-2x^2-4x leaves x^2+x+3; then 1(x^2-x-2) leaves 2x+5. Quotient 2x+1, remainder 2x+5. With x^2-x-2=(x-2)(x+1), 2x+5\equiv A(x+1)+B(x-2). At x=2: 9=3A, so A=3. At x=-1: 3=-3B, so B=-1. Answer: 2x+1+\frac3{x-2}-\frac1{x+1}. Check at x=0: \frac3{-2}=-1.5 and 1-1.5-1=-1.5.

7. 2x^2+7x+3: b^2-4ac=49-24=25=5^2, so it factorises: (2x+1)(x+3). Then x+8\equiv A(x+3)+B(2x+1). At x=-\frac12: \frac{15}2=\frac52A, so A=3. At x=-3: 5=-5B, so B=-1. Answer: \frac3{2x+1}-\frac1{x+3}.

8. x^2+4x+1: b^2-4ac=16-4=12, not a perfect square, so irreducible. 5x^2+15x+1\equiv A(x^2+4x+1)+(Bx+C)(x+2), so [x^2]: 5=A+B; [x]: 15=4A+2B+C; [\text{CT}]: 1=A+2C. At x=-2: 20-30+1=-9=A(4-8+1)=-3A, so A=3. Then B=2 from [x^2], C=-1 from [\text{CT}], and [x] checks: 12+4-1=15. Answer: \frac3{x+2}+\frac{2x-1}{x^2+4x+1}.

9. With a constant over x^2+1 (b^2-4ac=-4, irreducible): x^2+2x+3\equiv A(x^2+1)+C(x-1). At x=1: 6=2A, so A=3; but [x^2] needs 1=A. Contradiction, so the template fails: three coefficients to match, only two unknowns. With Bx+C: x^2+2x+3\equiv A(x^2+1)+(Bx+C)(x-1), so [x^2]: 1=A+B; [x]: 2=C-B; [\text{CT}]: 3=A-C. x=1 still gives A=3, so B=-2 and C=0, and [x] checks: 0+2=2. Answer: \frac3{x-1}-\frac{2x}{x^2+1}.

10. 8x+1\equiv A(4x+3)+B. [x]: 8=4A, so A=2. [\text{CT}]: 1=3A+B=6+B, so B=-5. Answer: \frac2{4x+3}-\frac5{(4x+3)^2}.

11. 12x^2+8x+5\equiv A(2x+1)^2+B(2x+1)+C=4Ax^2+(4A+2B)x+(A+B+C). [x^2]: 12=4A, so A=3. [x]: 8=12+2B, so B=-2. [\text{CT}]: 5=3-2+C, so C=4 (and x=-\frac12 gives 3-4+5=4=C directly). Answer: \frac3{2x+1}-\frac2{(2x+1)^2}+\frac4{(2x+1)^3}.

12. Nested, f(x)=[(x-2)x-5]x+6, and f(1)=[(-1)(1)-5](1)+6=0, so (x-1) is a factor. Dividing it out leaves x^2-x-6 (1+24=25, a perfect square), which is (x-3)(x+2). So x^3-2x^2-5x+6=(x-1)(x+2)(x-3). Then 7x^2-6x-25\equiv A(x+2)(x-3)+B(x-1)(x-3)+C(x-1)(x+2). At x=1: -24=-6A, so A=4. At x=-2: 28+12-25=15=15B, so B=1. At x=3: 63-18-25=20=10C, so C=2. Answer: \frac4{x-1}+\frac1{x+2}+\frac2{x-3}.