01 · Direct coefficient
Find the coefficient of x⁴ in (1 + 3x)⁷.
Hint
Set r = 4.
Worked solution
C(7,4)3⁴ = 35 × 81 = 2835
Understand · explore · practise
Finding binomial coefficients
Find a chosen coefficient without expanding everything. Use the general term, handle reciprocal powers, combine products and solve for unknown constants or exponents.
Before you startBinomial expansion, index laws and solving equations
01 / General term
The (r + 1)th term uses r copies of the second part.
In (a + b)ⁿ, the general term is C(n,r)aⁿ⁻ʳbʳ, for r = 0,…,n. Since r starts at zero, it is the (r + 1)th term in that order.
For (2 − x)⁶, r = 2 gives C(6,2)2⁴(−x)² = 240x². The term is 240x²; the coefficient of x² is 240.
Change the expression and r below. Powers inside a bracket affect the resulting power of x, so r need not equal that power.
Build one general termChoose and compare
C(6,2) × 2⁴ × (−x)²
= 240x²
In (2 − x)⁶, choosing r = 2 copies of −x gives C(6,2)2⁴(−x)² = 240x². The diagram shows one of the 15 choices of positions.
Watch the chosen factors form one general term
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / Choose a power
Write the power of x before doing the arithmetic.
In (a + bx)ⁿ, the x-power is r. To find xᵏ, set r = k. In an expression such as (a + bx²)ⁿ, the x-power is 2r instead.
Find the x⁴ coefficient in (2 − 3x²)⁵Worked example
C(5,r)2⁵⁻ʳ(−3x²)ʳ
The x-power is 2r.
2r = 4 ⇒ r = 2
This is the third term in this ordering.
C(5,2)2³(−3)² = 10 × 8 × 9 = 720
The coefficient is 720. There is no x³ term because 2r = 3 has no integer solution.
03 / Reciprocal powers
Combine the powers from both parts of the term.
For (x² − 1/x)⁶, with x ≠ 0, both factors in the general term contain x:
C(6,r)(x²)⁶⁻ʳ(−1/x)ʳ
= C(6,r)(−1)ʳx¹²⁻³ʳ
The constant term has exponent zero: 12 − 3r = 0, so r = 4. Its value is C(6,4)(−1)⁴ = 15.
If the required r is not an integer between 0 and n, that power is absent. A finite expansion can contain negative powers when the original bracket contains reciprocals; it is then not a polynomial in x.
A full reciprocal expansion
(x − 2/x)³
= x³ − 6x + 12/x − 8/x³, x ≠ 0
Retain the original domain even after collecting terms.
04 / Products of expansions
Collect every pair of terms whose powers add to the target.
To find xᵏ in a product, combine an xⁱ term from one factor with an xᵏ⁻ⁱ term from the other. Only expand as far as those contributions require.
(2 − x)⁶ = 64 − 192x + 240x² − 160x³ + …
For (3 + x)(2 − x)⁶, the x³ coefficient is 3(−160) + 240 = −240. Both the constant × x³ and x × x² contributions matter.
Two longer expansions
(1 + 2x)⁵ = 1 + 10x + 40x² + …
(2 − x)⁴ = 16 − 32x + 24x² + …
The x² coefficient in their product is 1 × 24 + 10 × (−32) + 40 × 16 = 344.
A missing term gives an equation
If (3 − px)(1 + x)⁵ has no x² term, its x² coefficient is 3C(5,2) − pC(5,1) = 30 − 5p. Set it equal to zero to obtain p = 6.
05 / Unknown constants
An even power can leave two possible constants.
If the coefficient of x² in (2 + cx)⁵ is 320, the coefficient formula gives:
C(5,2)2³c² = 320
80c² = 320 ⇒ c² = 4
c = 2 or −2
Keep both signs unless the question requires c to be positive. By contrast, a specified odd-power coefficient can determine the sign.
In (1 + cx)⁷, coefficient of x³ = −280
35c³ = −280 ⇒ c³ = −8 ⇒ c = −2
A coefficient is a signed number. Do not remove a minus sign just because the question calls it a coefficient.
06 / Unknown exponent
Use the factorial formula and enforce the integer condition.
Suppose n is a positive integer and the coefficient of x² in (1 − 2x)ⁿ is 84. A non-zero x² coefficient means n ≥ 2.
Solve for nWorked example
C(n,2)(−2)² = 84
C(n,2) = n(n − 1)/2.
2n(n − 1) = 84
n² − n − 42 = 0
Form an ordinary quadratic.
(n − 7)(n + 6) = 0
n = 7
Reject −6 because n is a positive integer.
Coefficient of x = −14
Coefficient of x³ = C(7,3)(−2)³ = −280
Use the permitted n to find the requested coefficients.
07 / Related coefficients
Translate each coefficient condition into algebra.
In (1 + px)⁶ with p ≠ 0, suppose the x coefficient is −q and the x² coefficient is 3q.
6p = −q, 15p² = 3q
15p² = −18p
p(15p + 18) = 0
p = −6/5, q = 36/5
The p = 0 branch is excluded by the assumption. Without that assumption, p = q = 0 would also satisfy the conditions.
A ratio of neighbouring coefficients
C(n,r + 1)/C(n,r) = (n − r)/(r + 1)
Cancel factorials to derive this ratio. For (1 + x)¹⁸, the x⁷ coefficient divided by the x⁶ coefficient is (18 − 6)/7 = 12/7. For (a + bx)ⁿ with a ≠ 0, the ratio also includes b/a.
State assumptions before dividing coefficients
For (A + x)ⁿ with A ≠ 0 and integer n ≥ 3, equality of the x² and x³ coefficients gives (n − 2)/(3A) = 1, so n = 3A + 2.
For A = 2, n = 8 and the first four terms are 256 + 1024x + 1792x² + 1792x³. If A = 0 or the relevant powers are absent, the ratio step may divide by zero; the conclusion need not follow. For instance x⁴ has both coefficients zero.
08 / Your turn
Identify the relevant term before calculating its coefficient.
State the coefficient alone when asked for a coefficient. Keep x ≠ 0 for reciprocal expressions.
02 · A negative coefficient
Find the coefficient of x³ in (3 − x)⁶.
Hint
The coefficient includes (−1)³.
Worked solution
C(6,3)3³(−1)³ = 20 × 27 × (−1) = −540
03 · Missing powers
Find the coefficients of x⁴ and x³ in (2 − 3x²)⁵.
Hint
The power is 2r.
Worked solution
For x⁴, r = 2 gives 720. For x³, r = 3/2 is not an integer, so the coefficient is zero.
04 · Constant term
Find the constant term in (2x − 1/x)⁴.
Hint
The x-power is 4 − 2r.
Worked solution
4 − 2r = 0 ⇒ r = 2
C(4,2)2²(−1)² = 24, x ≠ 0
05 · A product
Find the x² coefficient in (1 + 2x)⁵(2 − x)⁴.
Hint
Use power pairs (0,2), (1,1) and (2,0).
Worked solution
1 × 24 + 10 × (−32) + 40 × 16
= 344
06 · Two possible constants
The x² coefficient in (2 + cx)⁵ is 320. Find all real c.
Hint
80c² = 320.
Worked solution
c² = 4 ⇒ c = ±2
Both values meet the condition because the coefficient uses c².
07 · Odd power
The x³ coefficient in (1 + cx)⁷ is −280. Find c.
Hint
35c³ = −280.
Worked solution
c³ = −8 ⇒ c = −2
08 · Unknown exponent
For positive integer n, (1 − 2x)ⁿ has x² coefficient 84. Find n and its x³ coefficient.
Hint
C(n,2) = n(n − 1)/2.
Worked solution
2n(n − 1) = 84
(n − 7)(n + 6) = 0
n = 7; x³ coefficient = −280.
09 · Recover two coefficients
(1 + px)⁹ begins 1 + 27x + qx². Find p and q.
Hint
9p = 27.
Worked solution
p = 3
q = C(9,2)p² = 36 × 9 = 324
10 · Vanishing term
Find p so (3 − px)(1 + x)⁵ has no x² term.
Hint
Add the two contributions to x² and set their sum to zero.
Worked solution
3 × 10 − p × 5 = 0 ⇒ p = 6
11 · Adjacent coefficients
In (1 + x)¹⁷, find the x⁶ coefficient divided by the x⁵ coefficient.
Hint
Cancel C(17,6)/C(17,5).
Worked solution
(17 − 5)/6 = 2
12 · Two powers in the bracket
Find the x⁴ coefficient and the constant term in (2/x + x²)⁸.
Hint
The general x-power is −8 + 3r.
Worked solution
−8 + 3r = 4 ⇒ r = 4
x⁴ coefficient = C(8,4)2⁴ = 1120
A constant would require r = 8/3, not an integer. Its coefficient is zero. The expression requires x ≠ 0.
09 / Recap
Match the target power, then evaluate the coefficient.
- Tᵣ₊₁ = C(n,r)aⁿ⁻ʳbʳ, with r = 0,…,n.
- A coefficient excludes the variable power.
- Combine powers from both parts, especially with reciprocals.
- A product can contribute to one power in several ways.
- Retain both signs when an even power permits them.
- Check non-zero assumptions and the positive-integer exponent condition.
Section 1 of 9 · General term