To find an unknown constant in a binomial bracket, write the coefficient you are told about using the general term, put it equal to the given value, and solve. The constant usually appears as a power, so the equation is often a simple power equation such as k² = 9 or k³ = 8.
This builds directly on finding a specified term, and it is a common reason for the phrase “find the value of k” in a binomial question.
What is the method?
- Identify the required term from the power of x stated in the question.
- Find r using the power of the part that carries x.
- Write the coefficient using nCr and every numerical factor, including the powers of the constant part.
- Form the equation by setting this equal to the given value.
- Solve, then check each solution against any condition in the question.
Worked example
The coefficient of x³ in the expansion of (2 + kx)⁵ is 320. Find the value of k.
Step 1, general term: 5Cr × 2⁵⁻ʳ × (kx)ʳ.
Step 2, find r: the power of x is r, so r = 3.
Step 3, coefficient: 5C3 × 2² × k³ = 10 × 4 × k³ = 40k³.
Step 4, equation: 40k³ = 320.
Step 5, solve: k³ = 8, so k = 2.
Check: with k = 2, the coefficient of x³ is 40 × 8 = 320, which matches the question.
A second pattern: two coefficients equal
Suppose the coefficients of x² and x³ in (1 + ax)⁶ are equal and a ≠ 0. The coefficient of x² is 6C2 a² = 15a². The coefficient of x³ is 6C3 a³ = 20a³.
Setting them equal gives 15a² = 20a³, so a²(20a − 15) = 0. As a ≠ 0, a = 15/20 = 3/4.
The mistake to watch for
A frequent slip is to leave out the numerical part of the bracket when forming the equation.
Mistaken working: 5C3 × k³ = 320, so 10k³ = 320, k³ = 32
The student forgot the factor 2² = 4 that comes from the constant 2 in the bracket. The true coefficient is 40k³, not 10k³.
The correction is to write every factor of the term before simplifying: nCr, the power of the number, and the power of the letter. Check the result by substituting back: if k³ = 32 did not produce a whole-number cube, that is a clue that a factor is missing.
Check yourself
Try these, then open each answer.
1. The coefficient of x² in (1 + kx)⁴ is 54 and k > 0. Find k.
Show answer
4C2 k² = 6k² = 54, so k² = 9. Since k > 0, k = 3.
2. In the expansion of (2 + ax)⁴, the coefficients of x and x² are equal and a ≠ 0. Find a.
Show answer
Coefficient of x: 4C1 × 2³ × a = 4 × 8 × a = 32a. Coefficient of x²: 4C2 × 2² × a² = 6 × 4 × a² = 24a². Setting 32a = 24a² gives 24a² − 32a = 0, so 8a(3a − 4) = 0. As a ≠ 0, a = 4/3.
3. The coefficient of x⁴ in (x + k)⁶ is 240. Find the possible values of k.
Show answer
For x⁴, 6 − r = 4, so r = 2. The coefficient is 6C2 k² = 15k² = 240, so k² = 16. Then k = 4 or k = −4.
Where this leads next
Negative constants bring their own risk, so continue with tracking signs in an alternating expansion. When you are ready to mix all the skills, try the binomial expansion practice set.
Students who can solve the equation but struggle to write it often need someone to watch the set-up live. That is a regular part of online one-to-one Additional Mathematics tuition.