Why a Shortcut Has Conditions
A partial-fraction decomposition replaces one rational function with a sum of simpler rational functions. Finding its coefficients by expanding and comparing polynomial coefficients always works when the proposed form is correct, but it can create a system of equations. For certain terms, substitution makes one coefficient appear immediately. This shortcut is called the cover-up method.
The shortcut depends on having a simple linear factor: a factor that occurs exactly once in the denominator. Substitution at its root removes all the other terms from the relevant identity. Repeated factors behave differently, and an irreducible quadratic has no real root at which to perform the same substitution. In those cases, coefficient equations remain necessary.
The Cover-Up Formula
Suppose a proper rational function has denominator \(Q(x)=(x-a)S(x)\), where \(S(a)\ne 0\). Suppose its partial-fraction decomposition contains the term \(A/(x-a)\), and the other terms have denominators that do not vanish at \(x=a\). Multiplying the decomposition by \(Q(x)\) gives a polynomial identity. Substituting \(x=a\) makes every term other than the one containing \(A\) vanish.
Here \(P(x)\) is the numerator of the proper rational function, and the denominator factors are written in the stated form. This formula assumes that \(x-a\) occurs only once. If the original rational function is improper, first use polynomial division and apply the method to the proper remainder. If numerator and denominator have a common factor, cancel it before identifying the denominator’s factors.
Worked Example: Two Distinct Simple Linear Factors
Worked Example: Two Distinct Simple Linear Factors
Decompose \(\frac{3x+5}{(x-1)(x+2)}\). The denominator has two distinct simple linear factors, so write
Multiply by \((x-1)(x+2)\) to obtain the polynomial identity
To find \(A\), substitute \(x=1\). The left side is \(3(1)+5=8\). The right side is \(A(3)+B(0)=3A\), so \(A=8/3\).
To find \(B\), substitute \(x=-2\). The left side is \(3(-2)+5=-1\). The right side is \(A(0)+B(-3)=-3B\), so \(B=1/3\).
Thus the decomposition is
Check by recombining: the numerator on the right is \(\frac{8}{3}(x+2)+\frac{1}{3}(x-1)=\frac{8x+16+x-1}{3}=\frac{9x+15}{3}=3x+5\). The denominator is \((x-1)(x+2)\), as required.
Why Substitution Is Enough for a Simple Factor
The reason is that the multiplied expression is an identity, not merely an equation that happens to hold at a few values. For a simple factor \(x-a\), the term \(A/(x-a)\), after multiplication by the full denominator, contributes \(A S(x)\). At \(x=a\), this becomes \(A S(a)\). Every other partial-fraction term has a denominator that is nonzero at \(a\); after multiplication by the full denominator, its contribution retains the factor \(x-a\) and therefore becomes zero.
Since \(S(a)\ne 0\), division gives \(A=P(a)/S(a)\). This explains both how the shortcut works and why its conditions matter: the factor must occur only once, and the remaining denominator must not also vanish at the same value.
Repeated Linear Factors: Only One Coefficient Is Isolated
If the denominator contains \((x-a)^m\) for \(m>1\), the decomposition must include a term for each power from \(1\) through \(m\). Substitution at \(x=a\) does not determine all those coefficients. After multiplication by the common denominator, the term with \(1/(x-a)^m\) is the only one that remains nonzero at \(a\). Terms with lower powers retain at least one factor of \(x-a\) and vanish there.
Thus substitution may directly determine \(A_m\), the coefficient of the highest reciprocal power. The other coefficients require additional equations, usually obtained by equating coefficients in the polynomial identity or by using suitable substitutions after the identity has been formed.
Worked Example: A Repeated Linear Factor
Decompose \(\frac{2x+3}{(x-1)^2(x+2)}\). Since \(x-1\) is repeated, include both reciprocal powers:
Multiply by \((x-1)^2(x+2)\):
At \(x=1\), the left side is \(5\). The right side is \(3B\), so \(B=5/3\). This is the coefficient of \(1/(x-1)^2\), not the coefficient of \(1/(x-1)\).
To obtain the other coefficients, expand the identity. Since \((x-1)(x+2)=x^2+x-2\) and \((x-1)^2=x^2-2x+1\), the right side is
Compare coefficients with \(2x+3=0x^2+2x+3\):
Use \(B=5/3\). From \(A+C=0\), \(C=-A\). The second equation gives \(A+5/3+2A=2\), so \(3A=1/3\) and \(A=1/9\). Therefore \(C=-1/9\). The third equation checks: \(-2/9+10/3-1/9=3\).
The decomposition is
Check by recombining: its numerator over the common denominator is \(\frac{1}{9}(x-1)(x+2)+\frac{5}{3}(x+2)-\frac{1}{9}(x-1)^2\). The first and third terms combine to \(\frac{(x-1)[(x+2)-(x-1)]}{9}=\frac{x-1}{3}\). Adding \(\frac{5(x+2)}{3}\) gives \(\frac{x-1+5x+10}{3}=2x+3\). This recovers the original numerator.
Irreducible Quadratics: No Real Root to Cover
An irreducible quadratic is a quadratic polynomial that cannot be factored into real linear factors. Its partial-fraction numerator has degree at most one, so its general form is \((Bx+C)/(q(x))\), where \(q(x)\) is the irreducible quadratic. There is no real value \(x\) that makes \(q(x)=0\). Consequently, there is no real root to substitute that would eliminate the other terms and isolate \(B\) or \(C\). The coefficients must instead be determined from additional equations.
A simple linear factor elsewhere in the same denominator can still be handled by cover-up. It may give one coefficient quickly, but it does not supply both coefficients in the linear numerator over the quadratic.
Worked Example: A Simple Factor and an Irreducible Quadratic
Decompose \(\frac{3x^2-2x+3}{(x-1)(x^2+1)}\). The quadratic \(x^2+1\) is positive for all real \(x\), so it has no real root. Use the form
Multiply by the denominator:
The simple factor \(x-1\) permits cover-up: set \(x=1\). The left side is \(3-2+3=4\); the right side is \(2A\). Hence \(A=2\).
Expand the identity: \(A(x^2+1)+(Bx+C)(x-1)=(A+B)x^2+(C-B)x+(A-C)\). Equating the coefficients with \(3x^2-2x+3\) gives \(A+B=3\), \(C-B=-2\), and \(A-C=3\). Substituting \(A=2\), the first equation gives \(B=1\), and the third gives \(C=-1\). The middle equation checks because \(-1-1=-2\).
Therefore,
Check by recombining: the right-side numerator is \(2(x^2+1)+(x-1)^2=2x^2+2+x^2-2x+1=3x^2-2x+3\). The denominator is unchanged, confirming the decomposition.
A Common Mistake and Its Correction
A tempting but incorrect move is to treat a repeated factor as if its lowest-power coefficient were isolated by substitution. For example, in the repeated-factor identity above, setting \(x=1\) gives
not \(5=3A\). The \(A\)-term contains the factor \(x-1\) after multiplication by the common denominator, so it vanishes at \(x=1\). The \(B\)-term does not vanish. The correction is to inspect the multiplied identity before identifying which coefficient survives. For a factor repeated \(m\) times, direct substitution isolates the coefficient of \(1/(x-a)^m\), while the remaining coefficients require further equations.
Another limitation is practical: cover-up does not determine the two unknown constants \(B\) and \(C\) in a linear numerator over an irreducible quadratic. For that term, use the polynomial identity and equate coefficients.
Choosing the Right Method
| Denominator factor | What substitution can determine | What else is needed |
|---|---|---|
| A distinct simple linear factor \(x-a\) | The coefficient of \(1/(x-a)\) | Nothing for that coefficient; other unknowns may remain |
| A repeated linear factor \((x-a)^m\) | The coefficient of \(1/(x-a)^m\) | Additional equations for lower reciprocal powers |
| An irreducible quadratic | No coefficient by real-root cover-up | Equations for the linear numerator’s coefficients |
Cover-up is most useful as a selective shortcut. Apply it where a distinct simple linear factor provides a root, then use coefficient equations for what substitution cannot isolate. Finally, recombine the proposed terms to check that their numerator and denominator reproduce the original rational function.
Check Your Understanding
Answer each question by identifying both the available shortcut and its limits.
- For a proper rational function with denominator \((x-3)(x+1)\), what value of \(x\) isolates the coefficient of \(1/(x-3)\) after multiplying through by the common denominator?
- In a decomposition containing \(A/(x-2)+B/(x-2)^2\), which coefficient is isolated by substituting \(x=2\) into the multiplied identity?
- Why can’t real-root cover-up directly determine the numerator coefficients over \(x^2+4\)?
- What should be done before using cover-up if the rational function is improper?
- What algebraic check confirms that a proposed partial-fraction decomposition is correct?
