Decompose rational expressions into partial fractions

Compare polynomial coefficients

Expand, collect equal powers and solve for parameters after clearing a common denominator. Equality is required throughout the original domain, not merely at one convenient input

Factor and check the degrees first

A rational expression is proper when its numerator degree is smaller than its denominator degree

1. 1Divide if needed Divide an improper expression into a polynomial quotient and a proper remainder \(\displaystyle \begin{gathered}\frac ND=Q+\frac RD\\\deg R<\deg D\end{gathered}\)
2. 2Keep the original domain Factor the denominator over the real numbers and keep its real zeros excluded even after cancellation

Distinct linear factors need constant numerators

Use one constant numerator for each distinct linear factor, then match coefficients after clearing the denominator

1. 1Choose the pattern Keep the original denominator restrictions \(\displaystyle \begin{gathered}\frac{5x+1}{(x-1)(x+2)}\\=\frac{A}{x-1}+\frac{B}{x+2}\\x\ne1,-2\end{gathered}\)
2. 2Clear the denominator Multiply by both factors and collect like terms \(\displaystyle \begin{aligned}5x+1&=A(x+2)\\&\quad+B(x-1)\\&=(A+B)x\\&\quad+(2A-B)\end{aligned}\)
3. 3Match coefficients Match the coefficient of the linear term and the constant term to get simultaneous equations \(\displaystyle \begin{aligned}A+B&=5\\2A-B&=1\end{aligned}\)

Find the first coefficient by solving the simultaneous equations

Find the second coefficient by solving the simultaneous equations

Keep nonmonic factors exactly as written

Keep each linear factor as written so its scale is retained

1. 1Choose the pattern Use a constant numerator over each factor and retain both excluded inputs \(\displaystyle \begin{gathered}\frac{7x-1}{(2x-1)(x+1)}\\=\frac{A}{2x-1}+\frac{B}{x+1}\\x\ne\frac12,-1\end{gathered}\)
2. 2Clear the denominator Multiply each partial fraction by the full denominator \(\displaystyle \begin{aligned}7x-1&=A(x+1)\\&\quad+B(2x-1)\end{aligned}\)
3. 3Match coefficients Match the linear and constant terms \(\displaystyle \begin{aligned}A+2B&=7\\A-B&=-1\end{aligned}\)
4. 4Solve the equations Subtract the second equation from the first, then substitute back \(\displaystyle \begin{gathered}3B=8\\B=\frac83\\A=B-1=\frac53\end{gathered}\)

Enter A for that nonmonic-factor example

Include every power of a repeated linear factor

Include a separate numerator for each power of the repeated factor

1. 1Choose the complete pattern Keep both powers of the repeated factor and the original exclusions \(\displaystyle \begin{gathered}\frac{4x}{(x-1)^2(x+1)}\\=\frac{A}{x-1}+\frac{B}{(x-1)^2}\\\quad+\frac{C}{x+1}\\x\ne1,-1\end{gathered}\)
2. 2Clear the denominator Each term keeps the factors not cancelled by its own denominator \(\displaystyle \begin{aligned}4x&=A(x^2-1)\\&\quad+B(x+1)\\&\quad+C(x-1)^2\end{aligned}\)
3. 3Match all coefficients Match the quadratic, linear and constant terms in order \(\displaystyle \begin{aligned}A+C&=0\\B-2C&=4\\-A+B+C&=0\end{aligned}\)

Use the given substitution in the cleared identity to find the coefficient

Use the given coefficient in the linear equation to find the remaining coefficient

Three distinct factors give three constant numerators

A zero of a denominator factor isolates one coefficient in the cleared polynomial identity

1. 1Choose the pattern Use one constant numerator per distinct factor and keep all three exclusions \(\displaystyle \begin{gathered}\frac{2x^2-5x-4}{(x+1)x(x-2)}\\=\frac{A}{x+1}+\frac{B}{x}\\\quad+\frac{C}{x-2}\\x\ne-1,0,2\end{gathered}\)
2. 2Clear the denominator Use the polynomial identity, not the undefined original fractions, at the excluded inputs \(\displaystyle \begin{aligned}2x^2-5x-4&=Ax(x-2)\\&\quad+B(x+1)(x-2)\\&\quad+Cx(x+1)\end{aligned}\)
3. 3Isolate each coefficient Substitute the three factor zeros into the cleared identity \(\displaystyle \begin{aligned}x=-1:\quad3&=3A\\x=0:\quad-4&=-2B\\x=2:\quad-6&=6C\\A=1,\quad B&=2,\quad C=-1\end{aligned}\)
4. 4Check the expansion The recombined numerator agrees in every coefficient; the original exclusions remain \(\displaystyle \begin{gathered}x(x-2)+2(x+1)(x-2)\\-x(x+1)\\=2x^2-5x-4\end{gathered}\)

Use the given substitution in the cleared identity to find the coefficient

An irreducible real quadratic needs a linear numerator

Give an irreducible real quadratic a general linear numerator; a constant alone would miss valid decompositions

1. 1Choose the numerator degree The numerator degree must be less than the quadratic degree \(\displaystyle \text{Numerator: }Bx+C\)
2. 2Check positivity A completed square with a positive constant contributes no excluded real input \(\displaystyle \begin{gathered}(x-h)^2+k,\qquad k>0\\(x-h)^2+k>0\quad(x\in\mathbb R)\end{gathered}\)

Which pattern fits a proper fraction with the given denominator?

Worked linear–quadratic decomposition

Use a constant numerator over the linear factor and a linear numerator over the irreducible quadratic

1. 1Choose the pattern The quadratic is positive for real inputs, so only the linear factor excludes an input \(\displaystyle \begin{gathered}\frac{x^2+4x-1}{(x-1)(x^2+1)}\\=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}\\x\ne1\end{gathered}\)
2. 2Clear the denominator Expand both products before matching coefficients \(\displaystyle \begin{aligned}x^2+4x-1&=A(x^2+1)\\&\quad+(Bx+C)(x-1)\end{aligned}\)
3. 3Match coefficients Match the quadratic, linear and constant terms in order \(\displaystyle \begin{aligned}A+B&=1\\C-B&=4\\A-C&=-1\end{aligned}\)
4. 4Solve the equations Use the first and last equations in the middle equation, then substitute back \(\displaystyle \begin{gathered}B=1-A,\quad C=A+1\\(A+1)-(1-A)=4\\2A=4\\A=2,\quad B=-1,\quad C=3\end{gathered}\)

Enter B in that decomposition

Enter C in the same decomposition

Two irreducible quadratic factors need two linear numerators

Give each irreducible quadratic its own linear numerator

1. 1Choose the pattern Both quadratic factors are positive, so the real domain is unrestricted \(\displaystyle \begin{gathered}\frac{3x^2+x+5}{(x^2+1)(x^2+2)}\\=\frac{Ax+B}{x^2+1}\\\quad+\frac{Cx+D}{x^2+2}\\x\in\mathbb R\end{gathered}\)
2. 2Clear the denominator Multiply by both factors and collect equal powers \(\displaystyle \begin{gathered}3x^2+x+5\\=(Ax+B)(x^2+2)\\+(Cx+D)(x^2+1)\end{gathered}\)
3. 3Match coefficients Match the cubic, quadratic, linear and constant terms in order \(\displaystyle \begin{aligned}A+C&=0\\B+D&=3\\2A+C&=1\\2B+D&=5\end{aligned}\)
4. 4Solve the equations Subtract the first equation from the third and the second from the fourth, then substitute back \(\displaystyle \begin{aligned}A&=1& B&=2\\C&=-1& D&=1\end{aligned}\)

Enter D in that two-quadratic example

Divide before decomposing an improper fraction

Divide first, then decompose only the proper remainder

1. 1Start with the original expression Retain the original domain throughout the calculation \(\displaystyle \begin{gathered}\frac{x^4-3x^3+4x^2+x+1}{(x-1)(x^2+1)}\\x\ne1\end{gathered}\)
2. 2Expand and choose the first quotient term Name the numerator and denominator, expand the denominator, then divide the leading terms \(\displaystyle \begin{gathered}N=x^4-3x^3+4x^2+x+1\\D=(x-1)(x^2+1)\\=x^3-x^2+x-1\\\frac{x^4}{x^3}=x\end{gathered}\)
3. 3Subtract the first multiple Multiply the denominator by the first quotient term and subtract from the numerator \(\displaystyle \begin{gathered}xD=x^4-x^3+x^2-x\\N-xD\\=-2x^3+3x^2+2x+1\end{gathered}\)
4. 4Subtract the next multiple Divide the new leading terms, then subtract the next multiple of the denominator \(\displaystyle \begin{gathered}\frac{-2x^3}{x^3}=-2\\-2D=-2x^3+2x^2-2x+2\\(N-xD)-(-2D)\\=x^2+4x-1\end{gathered}\)
5. 5Read the quotient and remainder Stop because the remainder degree is less than the denominator degree \(\displaystyle \begin{gathered}Q=x-2\\R=x^2+4x-1\\\deg R=2<3=\deg D\end{gathered}\)
6. 6Clear the proper remainder denominator Use the complete linear–quadratic pattern and clear its denominator \(\displaystyle \begin{gathered}\frac{x^2+4x-1}{(x-1)(x^2+1)}\\=\frac A{x-1}+\frac{Bx+C}{x^2+1}\\x^2+4x-1=A(x^2+1)\\+(Bx+C)(x-1)\\=(A+B)x^2\\+(C-B)x+(A-C)\end{gathered}\)
7. 7Match and solve the coefficients Match powers, then use the first and last equations in the middle equation \(\displaystyle \begin{gathered}\begin{aligned}A+B&=1\\C-B&=4\\A-C&=-1\end{aligned}\\B=1-A,\quad C=A+1\\(A+1)-(1-A)=4\\2A=4\\A=2,\quad B=-1,\quad C=3\end{gathered}\)
8. 8Decompose the proper remainder Clear its denominator and match coefficients \(\displaystyle \begin{gathered}\frac{x^2+4x-1}{(x-1)(x^2+1)}\\=\frac2{x-1}+\frac{-x+3}{x^2+1}\end{gathered}\)
9. 9Keep the quotient in the full answer Add the polynomial quotient to the partial fractions and retain the excluded input \(\displaystyle \begin{gathered}x-2+\frac2{x-1}\\+\frac{-x+3}{x^2+1}\\x\ne1\end{gathered}\)

Write the polynomial quotient in the given form and enter its constant term

Repeat the full linear numerator at each quadratic power

A repeated quadratic requires a full linear numerator over each of its powers

1. 1Choose the complete pattern Do not omit the lower power; the quadratic is positive on the real line \(\displaystyle \begin{gathered}\frac{x^3+2x^2+4x+1}{(x^2+1)^2}\\=\frac{Ax+B}{x^2+1}\\\quad+\frac{Cx+D}{(x^2+1)^2}\\x\in\mathbb R\end{gathered}\)
2. 2Clear the denominator Only the lower-power fraction keeps a quadratic factor after cancellation \(\displaystyle \begin{gathered}x^3+2x^2+4x+1\\=(Ax+B)(x^2+1)\\+(Cx+D)\end{gathered}\)
3. 3Match coefficients Match the cubic, quadratic, linear and constant terms in order \(\displaystyle \begin{aligned}A&=1\\B&=2\\A+C&=4\\B+D&=1\end{aligned}\)
4. 4Solve the remaining coefficients Substitute the known cubic and quadratic coefficients \(\displaystyle \begin{aligned}C&=4-1=3\\D&=1-2=-1\end{aligned}\)

Enter C for that repeated-quadratic example

A zero coefficient does not restore an excluded input

A simpler formula may be defined at an input that the original expression excludes

1. 1Decompose on the original domain A zero coefficient does not remove the original denominator restriction \(\displaystyle \begin{gathered}\frac{2(x+2)}{(x-1)(x+2)}\\=\frac2{x-1}+\frac0{x+2}\\x\ne1,-2\end{gathered}\)
2. 2Simplify without filling the hole Equality to the original is asserted only on its original restricted domain \(\displaystyle \begin{gathered}\frac{2(x+2)}{(x-1)(x+2)}\\=\frac2{x-1}\\x\ne1,-2\end{gathered}\)
3. 3Distinguish the two domains The shorter formula has a value at this input, but the original expression is still undefined there \(\displaystyle \begin{gathered}x=-2\\\frac2{x-1}=\frac2{-3}=-\frac23\end{gathered}\)

After the displayed cancellation, may the specified input be included in the original expression’s domain?

Justify clearing denominators, ungraded

Explain why a cleared polynomial identity may be evaluated at a denominator zero without defining the original fraction there; verify the displayed equality in your ungraded written justification

Model: separate the polynomial statement from the rational domain

Multiplication by a nonzero common denominator is reversible on the original domain; matching every coefficient establishes a polynomial identity valid at all real inputs

1. 1Verify the polynomial identity The identity can be evaluated at the excluded inputs without defining the original fraction there \(\displaystyle 2(x+2)+3(x-1)=2x+4+3x-3=5x+1\)
2. 2Divide back only on the original domain Retain both exclusions when returning to the rational expression \(\displaystyle (x-1)(x+2)\ne0\iff x\ne1,-2\)

Prove uniqueness for distinct linear factors, ungraded

Prove that the displayed decomposition has at most one pair of coefficients for a fixed proper rational expression on its original domain; state exactly where distinctness is used in your ungraded proof

Model: subtract and evaluate the identity

Suppose two coefficient pairs represent the same fraction on the stated domain; their difference gives a polynomial zero at infinitely many inputs, hence an identity

1. 1Subtract the two representations Clear the common denominator on the original domain \(\displaystyle \begin{gathered}\frac A{x-r}+\frac B{x-s}=\frac{A^{\prime}}{x-r}+\frac{B^{\prime}}{x-s}\\r\ne s,\quad x\ne r,s\\(A-A^{\prime})(x-s)+(B-B^{\prime})(x-r)=0\end{gathered}\)
2. 2Evaluate the polynomial identity Distinct roots make each surviving factor nonzero and force the two coefficient pairs to agree \(\displaystyle \begin{gathered}x=r:\quad(A-A^{\prime})(r-s)=0\ \Longrightarrow\ A=A^{\prime}\\x=s:\quad(B-B^{\prime})(s-r)=0\ \Longrightarrow\ B=B^{\prime}\end{gathered}\)
3. 3Locate the distinctness requirement Coincident roots prevent these deductions and require the repeated-factor pattern instead \(\displaystyle r=s\ \Longrightarrow\ r-s=s-r=0\)

Verify with technology and report the domain, ungraded

Use symbolic technology to recombine both displayed answers, compare each numerator with its original and test an allowed rational input; explain how you retain original exclusions if a device simplifies a zero coefficient away

1. 1Linear–quadratic example Verify this decomposition on its stated domain \(\displaystyle \begin{gathered}\frac{x^2+4x-1}{(x-1)(x^2+1)}\\=\frac2{x-1}+\frac{-x+3}{x^2+1}\\x\ne1\end{gathered}\)
2. 2Improper example Retain the polynomial quotient as well as the partial fractions \(\displaystyle \begin{gathered}\frac{x^4-3x^3+4x^2+x+1}{(x-1)(x^2+1)}\\=x-2+\frac2{x-1}+\frac{-x+3}{x^2+1}\\x\ne1\end{gathered}\)

Model: identity verification is stronger than one numerical match

A numerical match is an error check; complete coefficient agreement proves the rational identity on its stated domain

1. 1Recombine the linear–quadratic answer The common numerator agrees in every coefficient \(\displaystyle \begin{gathered}\begin{gathered}\frac{x^2+4x-1}{(x-1)(x^2+1)}\\=\frac2{x-1}+\frac{-x+3}{x^2+1}\\x\ne1\end{gathered}\\2(x^2+1)+(-x+3)(x-1)=x^2+4x-1\end{gathered}\)
2. 2Verify the improper division Multiply the quotient by the denominator and add the proper remainder \(\displaystyle \begin{gathered}\begin{gathered}\frac{x^4-3x^3+4x^2+x+1}{(x-1)(x^2+1)}\\=x-2+\frac2{x-1}+\frac{-x+3}{x^2+1}\\x\ne1\end{gathered}\\(x-2)(x^3-x^2+x-1)+(x^2+4x-1)\\=x^4-3x^3+4x^2+x+1\end{gathered}\)
3. 3Check an allowed input Both expressions exclude the same input; one numerical match alone does not prove an identity \(\displaystyle \begin{gathered}x=0:\quad\text{linear–quadratic value}=1\\\text{improper value}=-1,\qquad x\ne1\end{gathered}\)

Choose the pattern before solving

Factor over the real numbers, divide if needed, include every repeated-factor power, use a lower-degree numerator for each factor, clear denominators and solve all coefficient conditions. Finally recombine and report the original domain