When two binomial expressions are multiplied, the coefficient of x^r in their product comes from every pair of terms whose exponents add to r. You usually do not need to expand either expression completely. List the possible power pairs, calculate their coefficients, then add those contributions with the correct signs.
For example, the coefficient of x³ in (1 + x)^4(1 − 2x)^3 is 8. It comes from four pairings: x⁰×x³, x¹×x², x²×x¹ and x³×x⁰. Their contributions are −8, 48, −36 and 4. This guide explains the method, shows why subtraction of expansions differs from multiplication, and develops variations involving missing powers, parameter values and exact verification.
A product is different from a sum of expansions
In P(x) + Q(x), the coefficient of x³ is simply the x³ coefficient of P plus the x³ coefficient of Q. There is no interaction between different powers.
In P(x)Q(x), an x term in P can combine with an x² term in Q to produce x³. Likewise, a constant term can combine with an x³ term, and an x² term can combine with an x term.
Therefore extracting a target coefficient from a product requires accounting for cross-products. Adding only the two separate x³ coefficients will generally miss several contributions.
This is the conceptual starting point. The arithmetic can be straightforward once each permissible pairing is listed, but the structure of multiplication must be understood before a binomial coefficient formula is applied.
Why exponents are added during multiplication
The index rule x^i × x^j = x^(i+j) explains which terms contribute to a selected power of x.
To obtain x³, the exponent pairs with non-negative integers are (0,3), (1,2), (2,1) and (3,0). If one factor has degree below an index, that pairing contributes zero.
For x⁴, the possible pairs are (0,4), (1,3), (2,2), (3,1) and (4,0). The list grows as the requested target power increases, but the rule remains the same.
Writing these pairs before calculating numerical coefficients is a useful discipline. It prevents a student from overlooking a middle cross-product or accidentally multiplying terms whose exponents give the wrong power.
The general coefficient-convolution rule
Let P(x) = Σ a_i x^i and Q(x) = Σ b_j x^j, where the sums contain their applicable polynomial powers. The product is formed by multiplying every a_i x^i by every b_j x^j.
The coefficient c_r of x^r is the sum of a_i b_j over every pair i + j = r. Equivalently, c_r = Σ a_i b_(r−i), treating unavailable coefficients as zero.
This is sometimes called coefficient convolution. The name is less important than the rule: identify power pairs that add to the target and sum the corresponding products.
For school work, a four-row table is often enough. The mathematical formula explains why that table contains all required terms rather than merely some plausible ones.
Record the needed coefficients of the first binomial
Take P(x) = (1 + x)^4. Its coefficients from x⁰ to x⁴ are 1, 4, 6, 4 and 1.
These follow from C(4,r) for r = 0, 1, 2, 3, 4. We need only the first four for an x³ target: a₀ = 1, a₁ = 4, a₂ = 6 and a₃ = 4.
There is no need to multiply four brackets repeatedly. The binomial theorem supplies each coefficient directly, and the target-power pairings show which ones matter.
A small coefficient table also helps distinguish the coefficient of x³ from the fourth term. Because indexing begins with r = 0, the coefficient of x³ is C(4,3), not C(4,2).
Record the signed coefficients of the second binomial
For Q(x) = (1 − 2x)^3, the coefficients from x⁰ to x³ are 1, −6, 12 and −8.
They come from C(3,j)(−2)^j. Thus b₀ = 1, b₁ = −6, b₂ = 12 and b₃ = −8.
The signs alternate because odd powers of −2 are negative and even powers are positive. Do not remove the signs merely because the two binomial factors are joined by multiplication rather than subtraction.
The signed coefficient table is the source of every contribution. Once its entries are correct, the remaining computation is simply matching index pairs and adding their products.
A complete x-cubed coefficient calculation
We want the x³ coefficient of (1 + x)^4(1 − 2x)^3. Four power pairings can contribute.
For (0,3), the contribution is a₀b₃ = 1(−8) = −8. For (1,2), it is a₁b₂ = 4(12) = 48.
For (2,1), it is a₂b₁ = 6(−6) = −36. For (3,0), it is a₃b₀ = 4(1) = 4.
Add all four signed contributions: −8 + 48 − 36 + 4 = 8. Therefore the coefficient of x³ is 8, and the corresponding term is 8x³. No full seventh-degree product expansion was necessary.
Check a nearby power to confirm the indexing
Find the coefficient of x² in the same product. The allowable power pairings are (0,2), (1,1) and (2,0).
Their contributions are a₀b₂ = 12, a₁b₁ = 4(−6) = −24 and a₂b₀ = 6.
The total is 12 − 24 + 6 = −6. Thus the coefficient of x² is −6.
This nearby calculation checks that the power pairings and signed coefficients were identified correctly. If an answer gives −6 for x³ and 8 for x², the student may have shifted the target index rather than made a difficult algebra error.
A fourth-power check uses five pairings
For the coefficient of x⁴ in P(x)Q(x), use pairs (1,3), (2,2), (3,1) and (4,0), together with (0,4) if the second factor had such a term.
Here Q has degree three, so b₄ = 0. The nonzero contributions are 4(−8) = −32, 6(12) = 72, 4(−6) = −24 and 1(1) = 1.
Add them: −32 + 72 − 24 + 1 = 17. Thus the x⁴ coefficient is 17.
A binomial with a smaller degree does not create an x⁴ term of its own, but it can still help create x⁴ when multiplied by positive powers from the other factor. This is why the pairing method is more general than inspecting each binomial separately.
A concise table is safer than a long expansion
For the x³ target, a simple table can have four rows labelled i = 0, 1, 2 and 3. The other index is j = 3 − i.
Record a_i, b_j and their product in each row. The result is a short signed sum with the correct structure visible.
This format is particularly useful when the factors contain a negative x coefficient, as in (1 − 2x)^3, or when one factor has a large exponent.
The purpose of a table is not to add decoration. It gives a complete inventory of all possible power combinations and makes the first wrong sign or omitted pairing easier to find during checking.
Use the binomial theorem directly for any pair
For P(x) = (1 + x)^m and Q(x) = (1 + kx)^n, the coefficient of x^i in P is C(m,i), while the coefficient of x^j in Q is C(n,j)k^j.
Therefore the coefficient of x^r in their product is Σ C(m,i)C(n,r−i)k^(r−i), over valid indices.
In the worked example m = 4, n = 3, k = −2 and r = 3. The four terms of this sum give −8, 48, −36 and 4.
This formula is the systematic version of the coefficient table. Students need not memorise it as another unrelated theorem: it follows immediately from the ordinary binomial general term plus the exponent rule for multiplication.
A symmetric product can simplify before expansion
Consider (1 + x)^4(1 − x)^4. Instead of expanding both fourth powers, group the two bases: [(1 + x)(1 − x)]^4.
Since (1 + x)(1 − x) = 1 − x², the product becomes (1 − x²)^4.
Every nonconstant term now has an even power of x. The coefficients of x, x³, x⁵ and x⁷ are all zero.
The x⁴ term arises by selecting (−x²) twice, giving C(4,2)(−1)²x⁴ = 6x⁴. This example shows that algebraic simplification before coefficient extraction can be even more efficient than constructing a pairing table.
The x-squared coefficient of the symmetric product
For (1 + x)^4(1 − x)^4 = (1 − x²)^4, the x² term comes from selecting one factor −x².
Its coefficient is C(4,1)(−1) = −4. There is no need to calculate cross-products separately.
The x⁴ coefficient is 6, as shown earlier, and the x⁶ coefficient is C(4,3)(−1)³ = −4.
Thus the simplified product has the power pattern 1 − 4x² + 6x⁴ − 4x⁶ + x⁸. Odd powers vanish through symmetry, not because the two original factors were separately missing odd terms.
Recognising the factorisation identity can save significant time, especially when the exponents are equal.
Vandermonde’s identity appears when the bases match
For (1 + x)^4(1 + x)^3, combine the powers immediately: the result is (1 + x)^7.
The x³ coefficient is therefore C(7,3) = 35.
If we used the pairing method, we would obtain C(4,0)C(3,3) + C(4,1)C(3,2) + C(4,2)C(3,1) + C(4,3)C(3,0).
That is 1 + 12 + 18 + 4 = 35. The equality is a special case of Vandermonde’s identity and confirms the coefficient-convolution rule.
This method comparison teaches a useful habit: before multiplying expansions, ask whether the bases are identical or can be combined using a simple algebraic identity.
Different exponents do not mean adding coefficient arrays
The correct identity is (1 + x)^4(1 + x)^3 = (1 + x)^7, because powers with the same base multiply by adding exponents.
It would be wrong to add the coefficients of the two original polynomials and treat the result as their product.
For example, the constant term of the product is 1 × 1 = 1, not the sum 1 + 1 = 2. The x coefficient is 4 × 1 + 1 × 3 = 7, consistent with C(7,1) = 7.
Understanding these low-degree cases is a useful conceptual check before attempting a higher target power. The algebraic operation between expressions determines how their coefficients combine.
An unknown multiplier can remove an x term
Suppose F(x) = (1 + x)^4(1 + kx)^3, where k is real. Find k such that the coefficient of x is zero.
The x term can arise in two ways: the x coefficient 4 of the first factor times the constant 1 of the second, or the constant 1 of the first times the x coefficient 3k of the second.
Thus the combined coefficient is 4 + 3k. Set it equal to zero: 4 + 3k = 0, giving k = −4/3.
This is a coefficient condition, not a requirement that the entire product vanish. Only one selected power of x has been cancelled; the constant and higher-power terms generally remain.
Find the next coefficient after choosing the multiplier
Continue with F(x) = (1 + x)^4(1 + kx)^3 and k = −4/3. Find the coefficient of x².
The three contributions are C(4,2) = 6, the product of the two x coefficients 4(3k) = 12k, and C(3,2)k² = 3k².
Therefore the x² coefficient is 6 + 12k + 3k². Substitute k = −4/3 to get 6 − 16 + 3(16/9).
This simplifies to −10 + 16/3 = −14/3. The x term has been eliminated, but the x² term is still present and negative.
This illustrates how one parameter can be fixed by one coefficient requirement without causing unrelated powers to vanish.
An x-cubed condition may lead to a cubic parameter equation
For the same F(x) = (1 + x)^4(1 + kx)^3, the x³ coefficient receives contributions from four index pairings.
Using the general-term formula, it is k³ + 12k² + 18k + 4. Requiring this coefficient to equal a specified value creates an equation in k that may be cubic.
A student should not assume that every unknown coefficient problem reduces to a linear equation. The degree in k depends on which term is requested and how k enters the binomial.
The essential method remains identical: find all contributing power pairs, form the exact coefficient polynomial, then solve the parameter equation by an appropriate algebraic method. The subsequent parameter roots must be checked against any stated restrictions.
An outside x changes which internal coefficients matter
Consider G(x) = x(1 + x)^4(1 − 2x)^3. To find its x⁴ coefficient, first notice that the outside x contributes one power.
Therefore we need the x³ coefficient of the inside product, which was 8. The x⁴ coefficient of G is also 8.
If we instead asked for G’s x³ coefficient, we would use the x² coefficient of the inside product, which was −6.
This is a common examination variation. A student who uses the full target index directly inside the binomial product will be one power too high. Account for every outside monomial before listing the internal exponent pairs.
The constant term of a product is often simple
If two factors are ordinary polynomials containing only non-negative powers of x, the constant term of their product is simply the product of their constant terms.
For (1 + x)^4(1 − 2x)^3, each constant term is 1, so the product’s constant term is 1.
No positive power from one factor can combine with a non-negative power from the other to produce exponent zero unless both exponents are zero.
This shortcut changes when negative powers of x are present. Then positive and negative exponents can cancel, and several pairings may contribute to a constant term. Always check the kind of expressions being multiplied.
A reciprocal-power product can have several constant contributions
Consider (1 + x)^2(1 + 1/x)^2 for x ≠ 0. The first factor has powers x⁰, x¹ and x²; the second has x⁰, x^(−1) and x^(−2).
A constant can come from power pairings (0,0), (1,−1) or (2,−2). Their coefficient products are 1·1 = 1, 2·2 = 4 and 1·1 = 1.
Thus the constant term is 1 + 4 + 1 = 6. This can also be seen by rewriting the expression as (1 + x)^4/x², whose constant term comes from the x² coefficient C(4,2) = 6.
The example shows that convolution is fundamentally about exponents summing to the target, whether the exponents are positive, zero or negative.
The degree gives a useful outer bound
A polynomial of degree m multiplied by a nonzero polynomial of degree n has degree m + n. The highest-power coefficient is the product of the leading coefficients.
For (1 + x)^4(1 − 2x)^3, the total degree is seven. The x⁷ coefficient is 1 × (−2)³ = −8.
There is no x⁸ term, so its coefficient is zero. These facts can be checked without working through any middle contributions.
Knowing the degree prevents impossible pairing attempts and gives quick tests for a student’s expanded result. If a purported expansion contains x⁹, or claims a positive leading coefficient for this example, the answer has a structural error.
A quick audit for missing or duplicated pairings
When extracting x^r, list pairs (i, r − i) in ascending i. The lower index starts at zero and stops when either factor’s exponent limit is reached.
If the same pair is counted twice or a valid pair omitted, the total coefficient changes. For the x³ example, the complete list has four distinct pairs and their signed contributions total 8.
A useful check is to reverse the order of the factors and repeat the pairing list. The result should remain unchanged because polynomial multiplication is commutative.
Students do not need to perform the calculation twice in every test, but knowing that a reversed pairing must agree helps diagnose a suspicious coefficient.
A compact examination solution format
First identify the target exponent and any outside monomial factor. Next simplify matching bases or symmetry identities if they can reduce the product directly.
If no shorter algebraic identity applies, list all index pairs whose powers sum to the target. Calculate the general-term coefficient for each factor at those indices.
Multiply each paired coefficient, preserving signs, and sum the contributions. State the requested coefficient as a number or the requested term as that number multiplied by x^r.
Finally check the product’s degree and, when convenient, one nearby coefficient or a special-value identity. This is shorter and easier to audit than expanding two large binomials without a plan.
Practice questions with checked answers
Question one: find the x³ coefficient of (1 + x)^4(1 − 2x)^3. Answer: 8.
Question two: find the x² coefficient of the same product. Answer: −6.
Question three: find its x⁴ coefficient. Answer: 17.
Question four: find the x³ coefficient of (1 + x)^4(1 + x)^3. Answer: C(7,3) = 35.
Question five: find the x⁴ coefficient of (1 + x)^4(1 − x)^4. Answer: 6.
Question six: choose k so that the x coefficient of (1 + x)^4(1 + kx)^3 vanishes. Answer: k = −4/3.
Question seven: find the constant term of (1 + x)^2(1 + 1/x)^2. Answer: 6, for x ≠ 0.
Attempt each by identifying allowable exponent pairs before consulting the results.
Syllabus context and next reading
Binomial products connect expansion coefficients, polynomial multiplication, index laws and parameter reasoning. Convolution language and reciprocal-power products can be treated as enrichment if not covered in the learner’s assessed syllabus.
For Singapore, consult the appropriate SEAB resources for 2026 O-Level Additional Mathematics, 2027 SEC G3 or 2027 SEC G2.
Companion guides include How to Find a Missing Coefficient in Combined Binomial Expansions, How to Find the Constant Term When a Binomial Contains Reciprocal Powers and How to Find Even and Odd Binomial Coefficient Sums Without Expanding. Explore the Additional Mathematics archive or enquire about eduKate Singapore three-student tutorials near Sixth Avenue MRT.
The durable skill is recognising which term pairings make the requested power, then adding all of their signed contributions without missing any. Properly taught kids shine a bright light into the future.