Why Substituting Roots Helps
When a rational function has a denominator that splits into distinct linear factors, its partial-fraction form assigns one unknown coefficient to each factor. The usual challenge is finding those coefficients efficiently.
The cover-up method uses a simple feature of the cleared-denominator identity: at a root of one denominator factor, every term containing that factor becomes zero. Only the coefficient attached to that factor remains. This makes it possible to calculate coefficients one at a time.
The method is a way to find coefficients in a specified partial-fraction form. It does not replace checking the result: after calculating the coefficients, multiply through by the denominator and verify the resulting polynomial identity.
The Setting and the Formula
Suppose a proper rational function has denominator consisting of distinct linear factors. In the monic-factor case, write it as
where the numbers \(r_1,\dots,r_n\) are distinct. The partial-fraction form for these factors is
Here \(A_1,\dots,A_n\) are constants to be determined. The condition \(\deg P<n\) says that the rational function is proper. If a rational function is not proper, polynomial division is needed before using this form.
Multiply both sides by the common denominator. This gives the cleared-denominator identity
To find \(A_i\), substitute \(x=r_i\). Every term on the right except the \(i\)th contains the factor \(x-r_i\), so those terms vanish. The remaining term gives a direct formula.
If the denominator includes a constant factor \(c\), so that it is \(c\prod_{j=1}^{n}(x-r_j)\), the coefficient formula becomes
The constant factor must be included: it remains in the cleared-denominator identity and therefore affects each coefficient.
Why the Method Works
The justification is built into the cleared-denominator identity. The following argument derives the formula for an arbitrary index \(i\).
Assume the rational function has the stated sum of fractions, with one constant \(A_j\) for each distinct factor \(x-r_j\).
Multiplying by \(\prod_{j=1}^{n}(x-r_j)\) gives a polynomial identity in which the term containing \(A_j\) is multiplied by all factors except \(x-r_j\).
For every \(j\ne i\), the term containing \(A_j\) includes \(x-r_i\), so it evaluates to zero. The \(A_i\) term evaluates to \(A_i\prod_{j\ne i}(r_i-r_j)\).
The identity therefore reads \(P(r_i)=A_i\prod_{j\ne i}(r_i-r_j)\). Since the roots are distinct, none of the factors \(r_i-r_j\) is zero, so division gives the cover-up formula.
This argument explains both the shortcut and its condition. Distinct roots ensure that the surviving product is nonzero. The substitution is performed in the cleared identity, not in the original rational function, which is undefined at \(x=r_i\).
Worked Examples
Worked Example: Two Distinct Linear Factors
Decompose
The denominator roots are \(1\) and \(-2\), so write
After multiplying by \((x-1)(x+2)\), the identity is
Set \(x=1\). The term containing \(B\) vanishes:
Set \(x=-2\). The term containing \(A\) vanishes:
Thus the proposed decomposition is
To check it, combine the terms on the right over the common denominator:
The original rational function is recovered, so the decomposition checks.
Worked Example: Three Distinct Linear Factors
Decompose
The numerator has degree \(2\), less than the denominator's degree \(3\), so the function is proper. Use the form
Clearing denominators gives
At \(x=2\), only the term containing \(A\) survives:
At \(x=-1\), only the term containing \(B\) survives:
At \(x=3\), only the term containing \(C\) survives:
Therefore the proposed decomposition is
To verify it, combine the numerators after clearing the common denominator:
Expand each product:
Collect the coefficients of \(x^2\), \(x\), and the constant:
The cleared numerators agree, verifying the decomposition.
Worked Example: A Coefficient Can Be Zero
Decompose
Use the form
After clearing denominators, evaluate at each root. At \(x=1\), the numerator itself is zero:
At \(x=-2\):
At \(x=-3\):
The decomposition is therefore
To check, put the right-hand side over the common denominator:
The numerator simplifies as follows:
This reproduces the original numerator. A zero coefficient is a valid outcome: the corresponding fraction simply contributes nothing.
Common Pitfalls and Limits
A frequent error is to evaluate the original rational function at a denominator root. For example, in the first worked example, substituting \(x=1\) into \(\frac{2x+5}{(x-1)(x+2)}\) is not permitted, because the denominator is zero. The method instead uses the cleared identity \(2x+5=A(x+2)+B(x-1)\), which is a polynomial identity. Substituting \(x=1\) there gives \(7=3A\), so \(A=7/3\).
Signs also require care. In that same calculation, the other factor at \(x=1\) is \(1+2=3\), so \(7=3A\), not \(7=-3A\). At \(x=-2\), the surviving factor is \(-2-1=-3\), giving \(1=-3B\) and hence \(B=-1/3\). Substitute the root into each remaining factor before simplifying.
The method described here applies directly to a proper rational function whose denominator is a product of distinct linear factors. If a factor is repeated, evaluating at its root makes multiple terms in the cleared identity vanish, and this shortcut alone does not isolate all the coefficients. Likewise, an irreducible quadratic factor is not a linear factor with a real root to substitute. Those cases require different coefficient procedures.
It is also important to keep the constant multiplier in the denominator. For example, if the denominator is \(2(x-1)(x+2)\), the factor \(2\) remains when clearing denominators. Omitting it changes the coefficient calculations. The safest practice is to write the full cleared-denominator identity before substituting any roots.
Why Verification Matters
Root substitution produces candidate coefficients efficiently, but a sign error or arithmetic slip can still occur. Multiplying the proposed decomposition by the common denominator checks the entire result at once: the resulting numerator must equal the original numerator as a polynomial.
For a denominator with distinct factors, the practical sequence is concise:
- Write the sum of simple fractions with one constant numerator per factor.
- Multiply through by the full denominator to obtain a polynomial identity.
- Substitute each factor's root into that identity and solve for its coefficient.
- Substitute the calculated coefficients into the proposed decomposition.
- Multiply back and confirm that the numerator simplifies to the original one.
The key idea is not that the original fraction can be evaluated at its poles. It cannot. Rather, clearing the denominator produces an identity that can be evaluated at those numbers, where the factors selectively vanish.
Check Your Understanding
For each question, focus on the cleared-denominator identity and the conditions for using root substitution.
- For \(\frac{3x+1}{(x-2)(x+1)}\), what identity should be formed before calculating either coefficient?
- In a denominator \((x-r_1)(x-r_2)(x-r_3)\), why do the terms other than the \(A_2\) term vanish when \(x=r_2\)?
- What condition on the roots ensures the product in the cover-up formula is nonzero?
- If the numerator evaluates to zero at one denominator root, what does that imply about the coefficient attached to that factor?
- Why is substituting a denominator root into the original rational function different from substituting it into the cleared-denominator identity?
