Reverses the addition of fractions, breaking a rational expression into a sum of simpler ones. Chooses the correct decomposition form from the denominator's factorisation, then finds the unknown numerators either by substituting convenient values or by matching coefficients.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 9 — Systems of Equations and Inequalities
§9.4 Partial Fractions, pp. 1104-1115
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §9.4 Partial Fractions §9.4, pp. 1104-1115 — the pages these objectives are drawn from
Warm-up
A familiar operation, looked at from the far end.
Discussion prompt
Add two over x minus one to three over x plus two. What does the result look like?
Hint: What is the common denominator?
Answer:
The common denominator is the product of the two, and the numerators combine into a single linear expression. The result is one fraction with a quadratic denominator.
Now run that backwards: given the single fraction, could you recover the two pieces? That is the question this section answers.
It matters because in calculus the separate pieces integrate easily and the combined fraction does not. Undoing the addition is what makes the integral possible.
Concept
A rational expression whose denominator factors can be written as a sum of simpler fractions, one for each factor, with numerators to be determined.
partial fraction decomposition — a rewriting of one rational expression as a sum of fractions whose denominators are the factors of the original denominator
\[ \frac{5x+1}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2} \]
The letters are placeholders for numbers not yet known. Writing the form correctly is the first half of the work, and solving for the letters is the second — and the second half cannot succeed if the first is wrong.
Figure (svg): A diagram showing addition of fractions in one direction and partial fraction decomposition as the reverse
OpenStax, Precalculus, §9.4 Partial Fractions §9.4, pp. 1104-1107
Section
Section 1
Concept
Decomposition applies only when the numerator's degree is less than the denominator's. If it is not, divide first and decompose the remainder.
Skipping this check produces a system with no solution, because the assumed form simply cannot represent an expression of that degree. The check costs one glance and prevents a page of futile algebra.
Figure (svg): A card stating that the numerator of each partial fraction has degree one less than its denominator
OpenStax, Precalculus, §9.4 Partial Fractions §9.4, pp. 1104-1108
Picture it
One rule generating every form in the catalogue.
Figure (svg): A card stating that the numerator of each partial fraction has degree one less than its denominator
The same rule governs the overall degree condition and each individual term's numerator. Understanding it once removes the need to memorise the four cases separately.
Worked example
The degree condition holds.
\[ \text{Can } \frac{5x+1}{x^2+x-2} \text{ be decomposed directly?} \]
Find the numerator's degree
Why: Linear.
\[ 1 \]
Find the denominator's degree
Why: Quadratic.
\[ 2 \]
Compare
Why: One is less than two.
Factor the denominator
Why: Ready to decompose.
\[ (x - 1) (x + 2) \]
Figure (svg): A card stating that the numerator of each partial fraction has degree one less than its denominator
\[ \frac{5x+1}{(x-1)(x+2)} \]
Verify: confirm the factorisation
Why: Expanding (x-1)(x+2) gives x squared plus x minus 2, matching the original denominator. Factoring is required before the form can be chosen, since the factors are what determine the terms.
OpenStax, Precalculus, §9.4 Partial Fractions §9.4, pp. 1105-1106
Prediction
The numerator is cubic and the denominator is quadratic.
Predict first
What must happen first?
Correct: Divide, then decompose the remainder.
Why: A decomposition can only produce numerators of degree below the denominator's, so a higher-degree numerator is outside its reach. Division extracts a polynomial part and leaves a remainder that satisfies the condition.
Worked example
The degree condition fails.
\[ \text{Prepare } \frac{x^2+3}{x^2-1} \text{ for decomposition.} \]
Compare the degrees
Why: Both quadratic.
Divide
Why: The quotient is one.
\[ 1\text{ with remainder } 4 \]
Write the result
Why: Quotient plus remainder fraction.
\[ 1 + \frac{4}{x ^{2} - 1} \]
Decompose the remainder only
Why: Now the degrees are right.
\[ \frac{4}{(x - 1) (x + 1)} \]
Figure (svg): The solution to Worked example divide first shown as a ladder of expressions, one row per legal move
\[ 1+\frac{4}{(x-1)(x+1)} \]
Verify: check by recombining
Why: One plus four over x squared minus one equals x squared minus one plus four, all over x squared minus one — which is x squared plus 3 over x squared minus 1, the original. The division was correct, and only the remainder needs decomposing.
OpenStax, Precalculus, §9.4 Partial Fractions §9.4, pp. 1106-1108
Trap
\[ \frac{x^2+3}{x^2-1}=\frac{A}{x-1}+\frac{B}{x+1} \]
Write the form for the denominator's factors
Why: The degree condition is not checked first.
Solving gives a system with no solution, because no constants can produce a quadratic numerator.
Compare the degrees first. A numerator of equal or higher degree needs division before anything else.
The assumed form can only produce numerators of degree less than the denominator's, so a top-heavy expression is outside its reach.
The system having no solution is the symptom, and the cure is a division step that takes one line.
Sorting
The numerator's degree must be lower.
Sort into buckets
Sort each expression.
Faded example
Comparing degrees before decomposing.
Fill in the blanks
\deg(\text1)=2 < \deg(\text___)=___ \;\Longrightarrow\; \text___
Why: A linear numerator over a quadratic denominator satisfies the condition, so the decomposition can be written directly. Equal degrees would require division first.
Step zero
You are given a rational expression to decompose.
Discussion prompt
What do you check before writing any form?
Hint: Two things, in order.
Answer:
Compare the degrees. If the numerator's is not lower, divide first and set the polynomial quotient aside.
Then factor the denominator completely, since the factors are what determine which form to write.
Both take seconds and both are prerequisites. Writing a form before factoring is guessing, and writing one before the degree check is guaranteed to produce an unsolvable system.
Section
Section 2
Concept
Each factor of the denominator contributes terms whose numerators have degree one less than that factor. A repeated factor contributes one term for each power up to its multiplicity.
The number of unknown letters always equals the denominator's degree, which is a useful check. If the form has more or fewer letters than that, something has been written wrongly.
| factor in the denominator | contributes |
|---|---|
| a distinct linear factor | one term, constant numerator |
| a linear factor repeated n times | n terms, one per power |
| a distinct irreducible quadratic | one term, linear numerator |
| a quadratic repeated n times | n terms, each with a linear numerator |
Figure (svg): Four cards giving the form of the decomposition for each type of factor in the denominator
OpenStax, Precalculus, §9.4 Partial Fractions §9.4, pp. 1107-1111
Picture it
Each card covers one kind of factor.
Figure (svg): Four cards giving the form of the decomposition for each type of factor in the denominator
The caption is the rule they all follow. Learning it saves memorising four patterns and handles combinations of them without further thought.
Worked example
One term per power.
\[ \text{Write the form for } \frac{3x-1}{(x-2)^2}. \]
Identify the factor
Why: Linear, repeated twice.
\[ (x - 2)\text{ squared} \]
Write one term per power
Why: First and second.
\[ \frac{A}{x - 2} + B / (x - 2) ^{2} \]
Check the numerators
Why: Linear factor, so constants.
Count the letters
Why: Two, matching the degree.
Figure (svg): Four cards giving the form of the decomposition for each type of factor in the denominator
\[ \frac{A}{x-2}+\frac{B}{(x-2)^2} \]
Verify: check the letter count
Why: The denominator has degree two and the form has two letters, which is the expected match. Writing only the squared term would give one letter, too few to represent a general linear numerator.
OpenStax, Precalculus, §9.4 Partial Fractions §9.4, pp. 1108-1109
Matching
Each kind of factor has one form.
Match the pairs
Why: The numerator's degree comes from the factor's degree and the number of terms comes from its multiplicity. Those two rules generate every case without any separate memorisation.
Worked example
A linear numerator is required.
\[ \text{Write the form for } \frac{2x^2+1}{(x-1)(x^2+4)}. \]
Identify the factors
Why: One linear, one quadratic.
Write the linear term
Why: Constant numerator.
\[ \frac{A}{x - 1} \]
Write the quadratic term
Why: Linear numerator.
\[ \frac{B x + C}{x ^{2} + 4} \]
Count the letters
Why: Three, matching the degree.
Figure (svg): The solution to Worked example an irreducible quadratic shown as a ladder of expressions, one row per legal move
\[ \frac{A}{x-1}+\frac{Bx+C}{x^2+4} \]
Verify: check the quadratic is irreducible
Why: x squared plus 4 has no real roots, so it cannot be factored further and takes a linear numerator as a single term. Had it factored, it would have split into two linear terms with constant numerators instead.
OpenStax, Precalculus, §9.4 Partial Fractions §9.4, pp. 1109-1111
Error analysis
A student writes the form for an expression with a quadratic factor.
Annotate
On: \( \frac{2x^2+1}{(x-1)(x^2+4)}=\frac{A}{x-1}+\frac{B}{x^2+4} \)
The letter count is the quickest check. It must equal the denominator's degree, and a shortfall means some numerator was written with too low a degree.
Prediction
The denominator has degree four.
Predict first
How many letters should the form contain?
Correct: Four.
Why: The number of unknowns always equals the denominator's degree, regardless of how it factors. That makes the letter count a quick check on whether the form was written correctly.
Faded example
A linear factor cubed.
Fill in the blanks
\frac23+\frac______}}}+\frac______}}}
Why: A factor repeated three times contributes three terms, one for each power from one up to the multiplicity. Three letters matches the denominator's degree of three.
Explain it to yourself
One rule generates all four forms.
Discussion prompt
State it and explain why it makes the catalogue unnecessary.
Hint: What determines a numerator's degree?
Answer:
Each numerator has degree one less than its denominator. A linear denominator takes a constant; a quadratic takes a linear expression.
Combined with one term per power of a repeated factor, that generates every form in the table without any of them being memorised separately.
And it explains the letter count: the total number of unknowns comes out equal to the denominator's degree every time. A rule that both generates the forms and checks them is worth more than a list of four cases.
Section
Section 3
Concept
After clearing denominators, substituting a root of one factor makes every other term vanish, isolating one unknown immediately.
This is fast but incomplete: it finds the unknowns attached to distinct linear factors and nothing else. Repeated and quadratic factors need coefficient matching for at least some of their letters.
Figure (svg): A contrast between the substitution shortcut and matching coefficients as methods for finding the unknown numerators
OpenStax, Precalculus, §9.4 Partial Fractions §9.4, pp. 1108-1112
Picture it
Substitution on the left, matching on the right.
Figure (svg): A contrast between the substitution shortcut and matching coefficients as methods for finding the unknown numerators
The left column's last row is its limitation. For a quadratic factor there is no real value making it zero, so no substitution isolates its numerator.
Worked example
Two substitutions, two answers.
\[ \text{Decompose } \frac{5x+1}{(x-1)(x+2)}. \]
Write the form
Why: One term per factor.
\[ \frac{A}{x - 1} + \frac{B}{x + 2} \]
Clear denominators
Why: Multiply through.
\[ 5 x + 1 = A(x + 2) + B(x - 1) \]
Substitute x = 1
Why: Kills the B term.
\[ 6 = 3 A, A = 2 \]
Substitute x = -2
Why: Kills the A term.
\[ -9 = -3 B, B = 3 \]
Figure (svg): A contrast between the substitution shortcut and matching coefficients as methods for finding the unknown numerators
\[ \frac{2}{x-1}+\frac{3}{x+2} \]
Verify: recombine the two fractions
Why: Two times (x+2) plus three times (x-1) is 2x plus 4 plus 3x minus 3, which is 5x plus 1 — the original numerator. Adding the pieces back is the definitive check and takes one line.
OpenStax, Precalculus, §9.4 Partial Fractions §9.4, pp. 1109-1111
Prediction
One factor is x minus three.
Predict first
Which value isolates its numerator?
Correct: Three, which makes that factor zero.
Why: Substituting a root of one factor makes every term still containing that factor vanish, leaving one unknown alone. The root of x minus three is three, not negative three.
Worked example
A repeated factor limits what it can find.
\[ \text{For } \frac{3x-1}{(x-2)^2}, \text{ what does substitution give?} \]
Clear denominators
Why: Multiply through.
\[ 3 x - 1 = A(x - 2) + B \]
Substitute x = 2
Why: Kills the A term.
\[ 5 = B \]
Note what is left
Why: A is not isolated by any substitution.
Find A by matching
Why: Compare the x coefficients.
\[ A = 3 \]
Figure (svg): The solution to Worked example substitution reaches only some letters shown as a ladder of expressions, one row per legal move
\[ \frac{3}{x-2}+\frac{5}{(x-2)^2} \]
Verify: recombine
Why: Three times (x-2) plus 5 is 3x minus 6 plus 5, which is 3x minus 1 — the original numerator. The substitution found one letter and coefficient matching found the other, which is the usual pattern with a repeated factor.
OpenStax, Precalculus, §9.4 Partial Fractions §9.4, pp. 1111-1112
Trap
\[ \text{substitute the roots, and all the letters will be found} \]
Rely on substitution alone
Why: Every unknown is assumed reachable by choosing the right value.
Letters attached to repeated or quadratic factors remain undetermined.
Substitution isolates only the numerators over distinct linear factors. A repeated factor's lower-power letters are not reachable.
An irreducible quadratic has no real root at all, so no substitution kills its term.
Use substitution for what it reaches and matching for the rest. Mixing the two methods in one problem is normal, not a sign of doing something wrong.
Faded example
After clearing denominators.
Fill in the blanks
5(1)+1=A(1+2)+B(1-1) \;\Longrightarrow\; 6=3A \;\Longrightarrow\; A=2
Why: Substituting the root of the second factor makes its term vanish, leaving one equation in one unknown. The arithmetic is immediate, which is what makes this method fast.
Sorting
It needs a real root of the factor.
Sort into buckets
Sort each factor.
Explain it
Choosing one value finds one unknown instantly.
Discussion prompt
Explain to a classmate why that value is chosen.
Hint: What happens to the other terms?
Answer:
After clearing denominators, each term carries the other factors as a product. Substituting a root of one factor makes those products zero.
So every term except one vanishes, leaving a single equation in a single unknown — which solves immediately with no system to handle.
It is a shortcut, not a different method: matching coefficients would give the same value with more work. A good explanation notes that the shortcut only exists where a real root does, which is exactly why quadratic factors fall outside it.
Section
Section 4
Concept
Expanding the cleared equation and collecting powers gives a system: the coefficient of each power on the left must equal the coefficient of that power on the right.
The method works because two polynomials are equal for every value only if their coefficients agree term by term. That is a genuine theorem rather than a convention, and it is what licenses the whole procedure.
Figure (svg): A contrast between the substitution shortcut and matching coefficients as methods for finding the unknown numerators
OpenStax, Precalculus, §9.4 Partial Fractions §9.4, pp. 1111-1114
Picture it
The right column works in every case.
Figure (svg): A contrast between the substitution shortcut and matching coefficients as methods for finding the unknown numerators
Its third row is the cost: a system rather than a single equation. For two or three unknowns that is manageable, and §9.6's matrix methods handle larger ones.
Worked example
Expand, collect, equate.
\[ \text{Decompose } \frac{2x+3}{(x+1)(x^2+1)}. \]
Write the form
Why: Constant over linear, linear over quadratic.
\[ \frac{A}{x + 1} + \frac{B x + C}{x ^{2} + 1} \]
Clear denominators
Why: Multiply through.
\[ 2 x + 3 = A(x ^{2} + 1) + (B x + C) (x + 1) \]
Expand and collect
Why: By powers.
\[ (A + B) x ^{2} + (B + C) x + (A + C) \]
Equate coefficients
Why: Three equations.
\[ A + B = 0, B + C = 2, A + C = 3 \]
Solve the system
Why: Three unknowns.
\[ A = \frac{1}{2}, B = -\frac{1}{2}, C = \frac{5}{2} \]
Figure (svg): A contrast between the substitution shortcut and matching coefficients as methods for finding the unknown numerators
\[ \frac{1/2}{x+1}+\frac{-\tfrac{1}{2}x+\tfrac{5}{2}}{x^2+1} \]
Verify: substitute a convenient value as a partial check
Why: At x equal to negative 1 the original gives 1 over 2, and the decomposition's first term is one half over zero — undefined, so instead check at x equal to 0: the original gives 3, and the pieces give one half plus five halves, which is 3. The values agree.
OpenStax, Precalculus, §9.4 Partial Fractions §9.4, pp. 1112-1114
Prediction
The right side is 2x plus 3, with no squared term.
Predict first
What is its squared coefficient?
Correct: Zero.
Why: A missing term means a coefficient of zero, not an absent equation. Writing the right side as zero times x squared plus 2x plus 3 makes the matching unambiguous.
Worked example
Substitute what you can, then match.
\[ \text{Decompose } \frac{x^2+2}{(x-1)(x^2+1)}. \]
Clear denominators
Why: Multiply through.
\[ x ^{2} + 2 = A(x ^{2} + 1) + (B x + C) (x - 1) \]
Substitute x = 1
Why: Isolates A immediately.
\[ 3 = 2 A, A = \frac{3}{2} \]
Match the squared coefficient
Why: A plus B equals one.
\[ B = -\frac{1}{2} \]
Match the constant
Why: A minus C equals two.
\[ C = -\frac{1}{2} \]
Figure (svg): The solution to Worked example combine both methods shown as a ladder of expressions, one row per legal move
\[ \frac{3/2}{x-1}+\frac{-\tfrac{1}{2}x-\tfrac{1}{2}}{x^2+1} \]
Verify: check at a convenient value
Why: At x equal to 0 the original gives negative 2, and the pieces give negative three halves plus negative one half, which is negative 2 — they agree. Substituting first reduced the system from three unknowns to two, which is the usual reason to combine the methods.
OpenStax, Precalculus, §9.4 Partial Fractions §9.4, pp. 1114-1115
Error analysis
A student matches terms from an expanded equation.
Annotate
On: \( (A+B)x^2+(B+C)x+(A+C)=2x+3 \;\Longrightarrow\; A+B=2 \)
A missing term on one side means a coefficient of zero, not a term to be skipped. Writing the right side with explicit zero coefficients before matching prevents the misalignment entirely.
Faded example
Equating the squared and linear coefficients.
Fill in the blanks
(A+B)x^2+(B+C)x+(A+C)=2x+3 \;\Longrightarrow\; A+B=0, \; B+C=2
Why: The right side has no squared term, giving a zero coefficient, and its linear coefficient is 2. Each power produces one equation, and together they form the system.
Sorting
Substitution reaches only some numerators.
Sort into buckets
Sort each situation.
Explain it to yourself
Equating coefficients is a legitimate step.
Discussion prompt
Explain what justifies it.
Hint: The equation holds for how many values?
Answer:
The cleared equation holds for every value of the variable, not just for some. It is an identity rather than an equation to solve.
And two polynomials that agree at every value must have identical coefficients — otherwise their difference would be a nonzero polynomial with infinitely many roots, which is impossible.
So matching coefficients is licensed by a genuine theorem. A good explanation notes the contrast with substitution, which uses only finitely many values and therefore reaches only some of the unknowns.
Section
Section 5
Concept
Adding the pieces back must reproduce the original expression exactly. The reason for wanting the pieces is that each is far easier to work with than the whole.
The recombination check is definitive and takes one line, which makes it worth doing every time. A value test is quicker but can pass by coincidence, so it is a partial check rather than a proof.
Figure (svg): A diagram showing addition of fractions in one direction and partial fraction decomposition as the reverse
OpenStax, Precalculus, §9.4 Partial Fractions §9.4, pp. 1104-1115
Picture it
The bottom line names the payoff.
Figure (svg): A diagram showing addition of fractions in one direction and partial fraction decomposition as the reverse
Recombining is also the check: going back to the right along the top arrow must return exactly the expression you started from.
Worked example
One line settles it.
\[ \text{Check that } \frac{2}{x-1}+\frac{3}{x+2}=\frac{5x+1}{(x-1)(x+2)}. \]
Use the common denominator
Why: The product of the two.
\[ (x - 1) (x + 2) \]
Scale each numerator
Why: By the other factor.
\[ 2(x + 2) + 3(x - 1) \]
Expand and collect
Why: Combine like terms.
\[ 2 x + 4 + 3 x - 3 \]
Compare
Why: It matches.
\[ 5 x + 1 \]
Figure (svg): A diagram showing addition of fractions in one direction and partial fraction decomposition as the reverse
\[ \text{verified} \]
Verify: note why this check is definitive
Why: Recombining reverses the decomposition exactly, so agreement proves the two expressions are identical rather than merely equal at some points. A single value test could pass by coincidence; this cannot.
OpenStax, Precalculus, §9.4 Partial Fractions §9.4, pp. 1110-1113
Prediction
You have found all the numerators.
Predict first
How do you prove the decomposition is correct?
Correct: Recombine and compare the numerator to the original.
Why: Recombining reverses the decomposition exactly, so matching numerators proves the two expressions are identical. A single value test can pass by coincidence and proves nothing.
Worked example
Each one is a standard integral.
\[ \text{Why decompose before integrating } \frac{5x+1}{(x-1)(x+2)}? \]
Consider the original
Why: No standard antiderivative.
Consider one piece
Why: A constant over a linear expression.
Consider the other
Why: The same shape.
Combine
Why: The sum of two logarithms.
Figure (svg): The solution to Worked example why the pieces are wanted shown as a ladder of expressions, one row per legal move
\[ 2\ln|x-1|+3\ln|x+2|+C \]
Verify: confirm the shapes
Why: A constant divided by a linear expression integrates to a constant times a logarithm, which is a standard result. Neither piece required anything beyond that, where the original expression had no directly applicable rule at all.
OpenStax, Precalculus, §9.4 Partial Fractions §9.4, pp. 1113-1115
Trap
\[ \text{both sides give }3\text{ at }x=0, \text{ so the decomposition is right} \]
Test one value and stop
Why: Agreement at a single point is taken as confirmation.
Two different expressions can agree at one point by coincidence.
Recombine the fractions. If the resulting numerator is identical to the original, the decomposition is proved.
A value test is a useful quick check that catches gross errors, but agreement at one point proves nothing in general.
Testing two or three values makes coincidence unlikely, but recombining makes it impossible — and it takes about the same time.
Faded example
Adding two partial fractions back together.
Fill in the blanks
2(x+2)+3(x-1)=5x+1
Why: Expanding gives 2x plus 4 plus 3x minus 3, which collects to 5x plus 1 — the original numerator. Matching exactly is what confirms the decomposition.
Sorting
Some checks prove and some only suggest.
Sort into buckets
Sort each check.
Explain it
Decomposing makes an expression longer, not shorter.
Discussion prompt
Explain to a classmate what it is for.
Hint: What comes next in calculus?
Answer:
The combined fraction is shorter but harder to work with. Each separate piece is a standard shape with a known antiderivative.
A constant over a linear factor integrates to a logarithm; a linear expression over an irreducible quadratic gives a logarithm and an inverse tangent. The combined fraction matches no standard form at all.
So the decomposition trades length for tractability. A good explanation notes that this is a recurring pattern — the useful form is often not the compact one, and the same technique reappears in differential equations and transform methods.
Comparison
Fill the blanks from memory. Which to use depends on the factors.
Comparison matrix
| substituting values | matching coefficients | |
|---|---|---|
| speed | very fast | slower, gives a system |
| distinct linear factors | finds every letter | also works |
| irreducible quadratics | cannot reach them | works |
| repeated factors | finds the top power only | finds all of them |
The two are complementary rather than competing. Substituting first and matching for what remains is the standard combination.
Pattern
Five steps, and the first two are prerequisites rather than work.
Step 3 is where the section's difficulty lies. A wrong form makes step 4 unsolvable, and the symptom — a system with no solution — points back here rather than to the arithmetic.
OpenStax Algebra and Trigonometry 2e, §11.4 Partial Fractions §11.4
Check
The degree condition.
Check your understanding
The numerator and denominator are both quadratic. What must happen first?
Answer: A
Why: A decomposition can only produce numerators of degree below the denominator's, so equal degrees are outside its reach. Division extracts a polynomial part and leaves a remainder that satisfies the condition.
Check
Choosing the form.
Check your understanding
What numerator goes over an irreducible quadratic factor?
Answer: A
Why: Each numerator has degree one less than its denominator, so a quadratic denominator takes a linear numerator. Using a constant gives too few unknowns to represent the general case.
Check
The methods.
Check your understanding
Which method can find every unknown in any decomposition?
Answer: A
Why: Matching works for every case because the cleared equation is an identity, so coefficients must agree power by power. Substitution reaches only the numerators over distinct linear factors, since only those have isolating real roots.
Real world
Control engineering decomposes transfer functions to read a system's behaviour.
Discussion prompt
An engineer decomposes a system's transfer function into partial fractions. What does each piece tell them?
Hint: What does each denominator factor correspond to physically?
Answer:
Each factor of the denominator corresponds to one mode of the system's response, and its root determines how that mode behaves over time.
A real root gives an exponential decay or growth; a complex pair — an irreducible quadratic — gives an oscillation. The numerator over each factor sets how strongly that mode is excited.
So decomposing converts one opaque expression into a list of behaviours with their strengths, which is exactly what a designer needs. The mathematics of this section is the standard first step in analysing any linear system, from circuits to aircraft controls.
Commit first
State your confidence along with your answer.
Predict first
Why does an irreducible quadratic factor need a linear numerator?
Correct: Each numerator has degree one less than its denominator.
Why: The rule is uniform across all four cases: a linear denominator takes a constant and a quadratic takes a linear expression. It also makes the total number of unknowns equal the denominator's degree, which is a useful check.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
A classmate's system for the numerators has no solution. What do you tell them to check?
Hint: The symptom points at an earlier step.
Answer:
Almost certainly the form was written wrongly, not the arithmetic. An unsolvable system is the standard symptom of a wrong decomposition form.
Two things to check: the degree condition — divide first if the numerator's degree is not lower — and the numerators, each of degree one below its denominator.
And the quick test: count the letters, which must equal the denominator's degree. A good explanation stresses that the symptom appears in step four but the cause is in step three, which is why re-checking the arithmetic is usually wasted effort.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The second is where the section's real difficulty lies, since a wrong form makes everything after it fail. The third is the shortcut most worth having automatic.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Write the numerator rule at the top, then derive all four decomposition forms from it. Underneath, decompose one expression with two distinct linear factors by substitution and one with a quadratic factor by matching. Finish by recombining one of them to verify.
If your four forms came out of the single rule rather than from memory, and your verification recombines rather than testing a value, the section's structure is on the page.
Recap
Five things, and the third is the one that decides whether the rest works.
| if you remember one thing | it should be this |
|---|---|
| about the form | each numerator's degree is one below its denominator's |
| about the letter count | it equals the denominator's degree, always |
| about substitution | it reaches distinct linear factors and nothing else |
| about checking | recombine; a single value test proves nothing |
Section 9.5 introduces matrices, which give the systems of this chapter a compact notation — and turn the elimination of §9.2 into a mechanical row procedure.
OpenStax, Precalculus, §9.4 Partial Fractions §9.4, pp. 1104-1115 — everything on these slides traces back here
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