Why Choose Equations Strategically?
A mixed partial-fraction decomposition can contain several unknown constants. Expanding every term and matching every power of \(x\) is valid, but it may create more algebra than necessary. A better first move is often to use the roots of the real linear factors: at such a root, many terms in the cleared identity vanish, leaving a direct equation for one coefficient.
After those substitutions, compare only the polynomial coefficients that are still needed. The aim is not to avoid coefficient matching entirely, but to use the simplest equations first and reduce the remaining system before expanding further.
The Identity to Solve
For a proper rational function with distinct real linear factors and an irreducible quadratic factor, the relevant decomposition has the form
where \(r_1\) and \(r_2\) are distinct real numbers, \(Q(x)\) is a quadratic with no real roots, and the numerator degree is smaller than the denominator degree. The linear numerator \(Cx+D\) is required for the quadratic term. The decomposition templates for these factors determine the form; here we focus on finding its unknown coefficients efficiently.
Multiplying both sides by the common denominator gives a polynomial identity:
Once denominators are cleared, this identity holds for every real \(x\). In particular, it holds at \(x=r_1\) and \(x=r_2\). At \(x=r_1\), the terms containing \(B\) and \(Cx+D\) vanish, leaving an equation for \(A\). At \(x=r_2\), the terms containing \(A\) and \(Cx+D\) vanish, leaving an equation for \(B\). The other coefficients can then be determined from the remaining polynomial identity.
Why the Root Substitutions Work
The method follows directly from the cleared identity. Suppose \(r_1\) and \(r_2\) are distinct and \(Q(r_1)\) and \(Q(r_2)\) are nonzero. Substituting \(x=r_1\) into the identity gives
Because \(r_1-r_2\ne 0\) and \(Q(r_1)\ne 0\), this equation determines \(A\). Substitution at \(x=r_2\) gives
which determines \(B\) under the same conditions. These are not shortcuts based on a pattern: they are consequences of evaluating a polynomial identity at particular values. The assumptions matter. If the linear factors coincide, the roots are not distinct; if a substituted value also makes another factor zero, the intended coefficient may not be isolated in this way.
After finding \(A\) and \(B\), subtract their contributions from the identity. The remaining expression is a polynomial identity involving only \(C\) and \(D\). Comparing coefficients then produces the equations needed for those constants.
A Practical Sequence
Use the appropriate numerator for each factor: a constant over a linear factor and a linear polynomial over an irreducible quadratic.
Keep each factor attached to the term it multiplies; this makes the useful substitutions visible.
Evaluate the whole cleared identity at each available real root, keeping careful track of signs and factor values.
After inserting the known constants, compare only the powers of \(x\) that determine the remaining unknowns.
Worked Examples
Worked Example: Two Real Roots and One Irreducible Quadratic
Decompose the rational function whose numerator is \(4x^3+12x^2-x-3\) and whose denominator is \((x-1)(x+2)(x^2+1)\). Use the form
Clear denominators:
At \(x=1\), the \(B\)-term and the quadratic-numerator term vanish. The left side is \(4+12-1-3=12\), and the remaining factor multiplying \(A\) is \((1+2)(1^2+1)=3\cdot2=6\). Thus \(12=6A\), so \(A=2\).
At \(x=-2\), the \(A\)-term and the quadratic-numerator term vanish. The left side is \(4(-8)+12(4)-(-2)-3=-32+48+2-3=15\). The remaining factor multiplying \(B\) is \((-2-1)((-2)^2+1)=(-3)(5)=-15\). Thus \(15=-15B\), so \(B=-1\).
Insert these values into the two known terms. Their sum is
Subtract this from the left side of the identity. The polynomial still to be supplied by \((Cx+D)(x-1)(x+2)\) is
Since \((x-1)(x+2)=x^2+x-2\), expand the remaining term:
Matching coefficients gives \(C=3\) from \(x^3\), and \(C+D=7\) from \(x^2\), so \(D=4\). Check the other two coefficients: \(D-2C=4-6=-2\), and \(-2D=-8\). Both agree with the remaining polynomial. Therefore the decomposition is
Worked Example: Track the Signs at Both Roots
Find the coefficients in
Clearing denominators gives
Set \(x=-1\). The \(B\)-term and the last term vanish. The left side is \(-2+5-3+30=30\). The factor multiplying \(A\) is \((-3)(5)=-15\). Hence \(30=-15A\), giving \(A=-2\).
Set \(x=2\). The \(A\)-term and the last term vanish. The left side is \(16+20+6+30=72\). The factor multiplying \(B\) is \((3)(8)=24\). Hence \(72=24B\), giving \(B=3\).
The known terms on the right expand to
Subtracting this from the left side leaves \(x^3-2x^2-x+2\). Since \((x+1)(x-2)=x^2-x-2\), expand:
Matching \(x^3\) gives \(C=1\). Matching \(x^2\) gives \(D-C=-2\), so \(D=-1\). The remaining coefficients check: \(-D-2C=1-2=-1\), and \(-2D=2\). Thus
Worked Example: One Real Root and a Quadratic
Find the coefficients in
After clearing denominators,
At \(x=2\), the term containing \(Bx+C\) vanishes. The left side is \(-4-6-5=-15\), while \(x^2+1=5\). Therefore \(-15=5A\), so \(A=-3\).
The known term is \(-3(x^2+1)=-3x^2-3\). Subtracting it from the left side leaves \(2x^2-3x-2\). Expand the remaining term:
Matching \(x^2\) gives \(B=2\). Matching \(x\) gives \(C-2B=-3\), so \(C-4=-3\) and \(C=1\). The constant term checks because \(-2C=-2\). The resulting decomposition is
A Common Sign Error
When substituting a root, evaluate the factor multiplying each coefficient as well as the numerator on the left. A sign error in one factor changes the coefficient equation.
For example, in the first worked example, at \(x=1\) the term containing \(B\) is \(B(x-1)(x^2+1)\), which vanishes. The term containing \(A\) is \(A(x+2)(x^2+1)=6A\). If a term \(B(x^2+1)\) is present in a different cleared identity, its value at \(x=1\) is \(2B\), not \(-2B\). In particular, the equation \(-2=2B\) gives \(B=-1\); writing \(-2=-2B\) would instead give \(B=1\). Evaluate each factor at the chosen root before simplifying.
When This Strategy Is Useful
Root substitutions are especially efficient when a cleared identity has distinct real linear factors and each substitution makes most terms vanish. Coefficient matching is then reserved for the terms that remain, rather than applied mechanically to the entire expanded identity from the start.
This method does not mean that every unknown is always determined by one root substitution. A real root isolates a coefficient only when the other terms vanish and the remaining multiplier is nonzero. If a factor is repeated, substituting its root may not provide enough independent equations to determine every associated coefficient. In that case, use the full polynomial identity and the appropriate equations for the repeated-factor terms.
Check Your Understanding
For each question, focus on which terms vanish and which equations remain.
- Why must the denominators be cleared before substituting a root into the identity?
- For a cleared identity with factors \((x-3)\), \((x+1)\), and \(x^2+4\), which terms vanish when \(x=3\)?
- In the identity \(P(x)=A(x+2)(x^2+1)+B(x-1)(x^2+1)+(Cx+D)(x-1)(x+2)\), what equation results from setting \(x=-2\)?
- After solving for the real linear-factor coefficients, why can matching only the remaining polynomial coefficients be more efficient?
- What must be checked before concluding that substitution at a real root isolates a coefficient?
