A binomial such as (2x − 1/x²)^6 contains both positive and negative powers of x. Its term independent of x comes from the expansion term whose combined exponent is zero. Use the binomial general term to set that exponent equal to zero; do not expand every term.
For (2x − 1/x²)^6, the general term indexed by r has x-power 6 − 3r. Setting 6 − 3r = 0 gives r = 2. The constant term is C(6, 2)2⁴(−1)² = 240. This guide shows how to locate the index, handle negative coefficients, account for outside powers, recognise when no constant term exists, and keep x ≠ 0 in the original domain.
A constant term has total x-power zero
An expression independent of x has x⁰ = 1 as its variable factor. A term such as 7x⁰ is therefore the constant 7. By contrast, 7x and 7/x are not constant terms.
When a binomial contains both x and 1/x², expanding its power produces terms with positive, zero and negative powers of x. The constant term appears precisely when these powers cancel in one expansion term.
This is an indexing question before it is an arithmetic question. A student should first calculate the power of x in the general term and solve for the term index that makes it zero.
Only after locating that index should the numerical binomial coefficient and other factors be evaluated. This order is far more efficient than writing an entire high-power expansion and searching for a variable-free term.
Why ordinary coefficient methods need a small extension
For (a + bx)^n, the general term normally contains a non-negative power x^r. In a reciprocal-power binomial, one part may contribute x^p while the other contributes x^(−q).
A typical expression is (a x^p + b x^(−q))^n for positive integers p and q, with x ≠ 0. Choosing r copies of the second term leaves n − r copies of the first.
Thus the combined exponent of x is p(n − r) − qr, not simply r. The constant-term condition is p(n − r) − qr = 0.
The general binomial theorem still determines the coefficient, but the power bookkeeping is more involved. Students who use r = 0 for every constant term will miss terms in which positive and negative powers cancel.
Derive the general term with reciprocal powers
For (a x^p + b x^(−q))^n, choose r copies of b x^(−q) and n − r copies of a x^p.
The resulting term is C(n, r)(a x^p)^(n−r)(b x^(−q))^r. Separate the constants from the powers of x.
Its coefficient is C(n, r)a^(n−r)b^r, and its x-power is p(n − r) − qr = pn − (p + q)r.
Therefore the constant-term index must satisfy r = pn/(p + q). This value must be an integer between 0 and n. If it is not, the finite binomial expansion has no x-independent term.
The complete opening example
Expand only as much as necessary to find the constant term of (2x − 1/x²)^6, for x ≠ 0.
The general term is C(6, r)(2x)^(6−r)(−x^(−2))^r. Its x-power is (6 − r) − 2r = 6 − 3r.
Set 6 − 3r = 0, giving r = 2. The numerical coefficient is C(6, 2)2^(6−2)(−1)².
Since C(6, 2) = 15, this is 15 × 16 × 1 = 240. The requested constant term is therefore 240, without expanding the remaining six terms. The sign is positive because the selected negative factor is squared.
Check the result by expanding nearby terms
The indexed terms of (2x − 1/x²)^6 begin with r = 0: 64x⁶. The term r = 1 is 6(2x)^5(−x^(−2)) = −192x³.
The term r = 2 is 15(2x)^4(x^(−4)) = 240x⁰ = 240. The next term r = 3 has x-power 6 − 9 = −3 and therefore includes x^(−3).
The powers proceed in steps of three: 6, 3, 0, −3, −6, −9 and −12. Exactly one term has exponent zero.
This nearby-term audit is useful because it confirms the indexing. It is not necessary to expand all seven terms on a school paper unless another part of the question explicitly asks for the full expression.
Why the exponent decreases in steps of three
Every time r increases by one, one factor 2x is replaced by a factor −1/x². The contribution to x-power changes from +1 to −2.
That replacement reduces the total exponent by three. Hence the sequence of powers is 6 − 3r for r = 0, 1, …, 6.
The constant term appears where this arithmetic progression of exponents reaches zero. In the worked example it occurs at r = 2.
This reasoning can be used before writing factorials. It also gives a quick prediction of whether a constant term is possible: zero must occur among the integer-indexed powers produced by replacing the factors one at a time.
An example with a positive reciprocal coefficient
Find the constant term of (2x + 3/x²)^6. The x-exponent is still 6 − 3r, so the constant index remains r = 2.
The coefficient is C(6, 2)2⁴3² = 15 × 16 × 9 = 2160.
The positive sign follows directly from the positive inner coefficients. The power selection is unchanged because the x-powers have not changed.
This variation separates two decisions. The powers of x determine which term can be constant, while the numerical coefficients determine the value and sign of that term. Students should make those decisions in that order.
A different ratio of positive and negative powers
Consider (3x² + 2/x)^6. Here the first factor contributes x² and the second contributes x^(−1).
The exponent in the general term is 2(6 − r) − r = 12 − 3r. Set it equal to zero, giving r = 4.
The numerical coefficient is C(6, 4)3^(6−4)2⁴ = 15 × 9 × 16 = 2160.
This coincidentally matches the numerical result in the previous example, but the constant comes from a different index. Similar-looking answers do not justify copying an index from one expression to another. Derive the exponent each time.
A negative coefficient can produce a negative constant term
Consider (2x − 1/x²)^3. The total x-exponent in the r-th term is 3 − 3r.
The constant index is r = 1. Its coefficient is C(3, 1)2^(3−1)(−1)^1 = 3 × 4 × (−1) = −12.
Thus the constant term is −12. The negative sign survives because only one copy of the negative reciprocal factor is selected.
This contrasts with the sixth-power example, where r = 2 and the negative factor was raised to an even power. Never assume the constant term has the same sign as the first binomial term; the sign depends on the selected index.
An example with exact fractional coefficients
Take (x² − 1/(2x))^6. The x-exponent is 12 − 3r, so r = 4.
The constant coefficient is C(6, 4)1²(−1/2)^4 = 15 × 1/16 = 15/16.
The exponent four makes the negative reciprocal coefficient positive. The fractional magnitude is retained exactly.
This example tests whether the student can distinguish 1/(2x) from 2/x. The constant attached to x^(−1) is 1/2, not 2, and raising it to the fourth power changes the answer substantially. Protective brackets are useful in the general term.
Some expressions have no constant term
Consider (x² + 1/x)^5. Its general x-exponent is 2(5 − r) − r = 10 − 3r.
To obtain a constant term, we would need 10 − 3r = 0, so r = 10/3. But r counts the number of selected factors and must be an integer from 0 to 5.
No such binomial term exists. Therefore the expansion has no term independent of x, and its constant coefficient is zero.
Do not round r to 3 or 4. The general term’s index is discrete. A fractional index here is evidence that no valid term has exponent zero, not an approximation to a suitable term.
Why an integer index is essential
In a finite expansion (A + B)^n with non-negative integer n, every term arises by selecting an integer number r of copies of B.
That count can never be fractional or exceed n. The valid indices are 0, 1, 2, …, n.
Thus r = pn/(p + q) is only a candidate until its integrality and range have been checked. If the value is an integer in that range, one selected term can be constant. Otherwise none can.
This is a mathematical admissibility condition similar to the non-negative restriction in quadratic substitutions. An algebraic equation for a new quantity is not the end of the solution until the meaning and allowed values of that quantity are respected.
An outside power changes the required exponent
Suppose the expression is x(2x − 1/x²)^5. The outside x contributes one additional power to every term.
Inside the binomial, the exponent is 5 − 3r. After multiplication by the outside x, the total exponent is 1 + 5 − 3r = 6 − 3r.
Set it equal to zero, giving r = 2. The numerical coefficient is C(5, 2)2^(5−2)(−1)² = 10 × 8 = 80.
Therefore the complete expression has constant term 80. Without accounting for the outside x, the student would solve 5 − 3r = 0 and incorrectly conclude that no constant term exists.
An outside factor may also eliminate the constant
Consider x³(2x − 1/x²)^4. Inside the fourth power, the x-exponent in the indexed term is 4 − 3r.
Including the outside x³ gives total exponent 7 − 3r. Requiring this to equal zero produces r = 7/3, which is not an integer.
Hence there is no constant term. This is true even though the inside binomial has multiple terms and the outside factor is a simple power of x.
The general lesson is to include every multiplicative power of x in the exponent equation before solving for r. The final total exponent, not the exponent inside one subexpression, decides whether the whole term is independent of x.
A general formula with an outside power
For x^s(a x^p + b x^(−q))^n, the r-th term has total power s + pn − (p + q)r.
A constant term requires r = (s + pn)/(p + q), with r a whole number between zero and n. The corresponding numerical coefficient is C(n, r)a^(n−r)b^r.
The formula assumes the expression is interpreted for x ≠ 0 where the reciprocal powers appear. It also assumes an ordinary finite binomial expansion with integer n ≥ 0.
A student does not need to memorise this extended result separately. It follows immediately by multiplying the outside x^s into the same general binomial term and collecting powers carefully.
A sum of two expansions can cancel constant terms
Suppose F(x) = (2x − 1/x²)^6 + k(3x² + 2/x)^6, with x ≠ 0. We have already calculated the first constant term as 240.
The second sixth power has constant term 2160. Therefore F has constant term 240 + 2160k.
If the question requires no term independent of x, set 240 + 2160k = 0, giving k = −240/2160 = −1/9.
Only the constant-term contributions were needed. The other terms can remain unexpanded because adding polynomials or Laurent polynomials combines coefficients of matching powers independently.
Why cancellation of constants does not cancel the whole expression
When k = −1/9 in the previous example, the constant coefficient is zero. That does not mean F(x) is identically zero.
The two expansions contain numerous nonconstant powers of x, and their coefficients generally do not cancel pairwise. The condition controls one coefficient only.
This distinction matters in many binomial parameter problems. “Find k so that the constant term vanishes” is an equation about a particular coefficient, not an instruction to set the entire function equal to zero.
A student who solves F(x) = 0 for x has changed the question. Instead, locate the relevant power in each expansion and equate their combined coefficient to the requested value.
Reciprocal powers impose an original domain
The expression (2x − 1/x²)^6 is undefined at x = 0 because 1/x² is undefined there. Expanding its powers and identifying a constant term does not restore x = 0 to the original domain.
The algebraic term 240 is valid as part of the expansion for nonzero x. It is not evidence that the entire original expression can be evaluated at zero.
This is a useful distinction between an algebraic coefficient and a function value. The constant term is independent of x, but the full expression may still contain positive and negative powers that prevent evaluation at x = 0.
Students should carry the restriction x ≠ 0 from the original reciprocal expression, especially if a later part asks for the function’s value or domain.
A coefficient of a nonzero power uses the same method
Suppose instead of the constant term, we want the coefficient of x³ in (2x − 1/x²)^6.
The x-exponent is 6 − 3r. Set it equal to 3, obtaining r = 1. The coefficient is C(6, 1)2⁵(−1) = 6 × 32 × (−1) = −192.
For the coefficient of x^(−3), set 6 − 3r = −3, giving r = 3. Its coefficient is C(6, 3)2³(−1)³ = 20 × 8 × (−1) = −160.
Thus the method is not limited to constant terms. Any requested power m is found by solving the exponent equation pn − (p + q)r = m, including outside powers where present.
Sign bookkeeping deserves its own check
The coefficient of the indexed term includes b^r. If b is negative, the coefficient is negative when r is odd and positive when r is even, assuming the other numerical factors are positive.
In the sixth-power opening example, r = 2 and the contribution is positive 240. In the third-power example, r = 1 and the contribution is negative 12.
Do not copy the sign of the original minus sign directly into the final answer without raising it to the selected index. Likewise, do not replace (−2)^3 by +8 merely because x-power cancellation has already been achieved.
A short sign audit after locating r is enough: identify the sign of each factor, its exponent, and any outside multiplier before multiplying the numerical coefficient.
The middle term is not necessarily constant
For an even binomial exponent, the middle term often has index r = n/2. But whether that term is independent of x depends on the powers of x in the two binomial components.
In (2x − 1/x²)^6, the constant occurs at r = 2, whereas the middle index for seven terms is r = 3. The middle term contains x^(−3), not x⁰.
The constant-term index is determined by balancing exponents, not by the visual location of the term in the expansion. Confusing the two concepts is a common shortcut error.
Use the equation for total x-power every time. If the answer happens to be the middle term in a particular expression, treat that as a consequence of its powers, not as a general rule.
A short practice set with checked answers
Question one: find the constant term in (2x − 1/x²)^6. Answer: 240.
Question two: find the constant term in (2x + 3/x²)^6. Answer: 2160.
Question three: find the constant term in (3x² + 2/x)^6. Answer: 2160.
Question four: find the constant term in (2x − 1/x²)^3. Answer: −12.
Question five: does (x² + 1/x)^5 contain a constant term? No; the required index is 10/3, not an integer.
Question six: find the constant term in x(2x − 1/x²)^5. Answer: 80.
Question seven: determine k so that (2x − 1/x²)^6 + k(3x² + 2/x)^6 has no constant term. Answer: k = −1/9.
For each, write the total exponent before calculating any binomial coefficients.
A concise examination solution format
Write the indexed general term C(n, r)A^(n−r)B^r, replacing A and B by the actual terms including powers of x.
Collect the exponent of x into a single expression in r. Set that exponent equal to zero for a constant term, or equal to the requested power for a different coefficient.
Solve for r and verify that it is an integer between zero and n. If it is not, state that no such term appears. If it is admissible, compute the numerical coefficient and carefully handle negative signs and fractions.
Finally include any outside x-power and original domain restriction. This sequence is typically much shorter and safer than expanding every term and searching through a long expression.
Scope, learning transfer and related reading
Constant terms in reciprocal-power binomials extend ordinary binomial coefficients into expressions with negative powers of x. Depending on the student’s course, this can be enrichment rather than a compulsory assessed problem, so match examples to the school syllabus and teacher’s instructions.
For Singapore, refer to the appropriate SEAB documents for 2026 O-Level Additional Mathematics 4049, 2027 SEC G3 K341 or 2027 SEC G2 K232 as relevant.
Related reading includes How to Find a Missing Coefficient in Combined Binomial Expansions, How to Recover a Binomial Exponent from Consecutive Coefficients and How to Track Domain Restrictions Through Composite Functions. Browse the Additional Mathematics archive or enquire about eduKate Singapore’s maximum three-student tutorials near Sixth Avenue MRT.
The main skill is not memorising one coefficient. It is identifying the exact indexed term whose variable powers cancel and checking that such a term actually exists. Properly taught kids shine a bright light into the future.