Adapt a binomial expansion and transform its validity interval
Factor out the constant, identify the small variable and translate its strict bound back to x.
Exam scope and assessment notes
Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027, section 3.1. Paper 1 knowledge is assumed. Shared modulus, polynomial division, factor and remainder lessons are linked from both sections 2.1 and 3.1. Paper 3 adds partial fractions and rational-power binomial expansions. Partial-fraction numerators may have the same degree as the denominator, but higher-degree numerators are excluded. General-term questions for the rational binomial expansion are not included. Use the stated strict validity condition. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Use rational binomial coefficients and fractional powers.
- Factor a constant from a bracket and solve a strict modulus inequality.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What factor is outside (1 − x/4)⁻¹ when (2 − x/2)⁻¹ is rewritten?
What you will learn
- Rewrite (a + bx)ⁿ in the form aⁿ(1 + u)ⁿ with a suitable real positive constant a.
- Include the outside factor in every coefficient and convert |u| < 1 into the correct interval for x.
Extract the constant before applying the standard series
For a positive real constant a, write (a + bx)ⁿ = aⁿ(1 + bx/a)ⁿ. The standard variable is u = bx/a, not x itself. Expand in u, substitute its expression and multiply every term by aⁿ. For fractional powers, this positive-constant form avoids silently choosing an invalid real branch; the supplied interval also keeps the bracket positive in these examples.
For (2 − x/2)⁻¹, factor 2 to get (1/2)(1 − x/4)⁻¹. Using u = −x/4 gives (1/2)[1 + x/4 + x²/16 + x³/64 + …] = 1/2 + x/8 + x²/32 + x³/128 + … . The outside factor is 1/2, not 2. Multiplying only the constant term by it would spoil the other coefficients.
Translate the bound and distinguish it from the function's domain
The condition is |−x/4| < 1, hence |x| < 4 or −4 < x < 4. It is not |x| < 1. The original reciprocal (2 − x/2)⁻¹ is undefined at x = 4 but is defined at x = −4; both endpoints are excluded from the strict binomial interval used here. Do not label every excluded series endpoint as a pole of the function.
For √(4 − x) = 2(1 − x/4)¹ᐟ², substitute u = −x/4 into 1 + u/2 − u²/8 + u³/16 + … . After multiplication by 2 the terms are 2 − x/4 − x²/64 − x³/512 + … , again for |x| < 4. The original real square root exists for x ≤ 4, which is a larger domain than this local expansion interval. Always state the transformed strict condition alongside the coefficients.
Pause and explain
For (2 − x/2)⁻¹, what does the binomial condition |−x/4| < 1 give?
Worked example
Expand (2 − x/2)⁻¹ through x³ and state its validity interval.
Show the worked solution
- Write (1/2)(1 − x/4)⁻¹ and choose u = −x/4.
- Expand (1 + u)⁻¹ = 1 − u + u² − u³ + … , then multiply by 1/2.
- The condition |u| < 1 becomes −4 < x < 4.
Answer 1/2 + x/8 + x²/32 + x³/128 + … , |x| < 4.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Expand (3 + 2x)⁻¹ through x³ and state its interval.
Give me a hint
Factor out 3, use u = 2x/3 and include the multiplier 1/3.
Compare my reasoning
- Rewrite the function as (1/3)(1 + 2x/3)⁻¹.
- Substitute u = 2x/3 into 1 − u + u² − u³ and multiply by 1/3.
- The result is 1/3 − 2x/9 + 4x²/27 − 8x³/81 + … , with |2x/3| < 1 giving |x| < 3/2.
1/3 − 2x/9 + 4x²/27 − 8x³/81 + … , |x| < 3/2.
Look for these in your work
- Scale every power and every coefficient.
- Transform the validity condition to x.
Expand √(4 − x) through x³ and compare the expansion interval with the original real domain.
Give me a hint
Use outside factor 2 and variable −x/4 in the square-root series.
Compare my reasoning
- Rewrite √(4 − x) as 2(1 − x/4)¹ᐟ².
- Substitute into 1 + u/2 − u²/8 + u³/16, obtaining 2 − x/4 − x²/64 − x³/512 + … .
- The expansion requires |x| < 4, whereas the original real square root is defined for x ≤ 4.
2 − x/4 − x²/64 − x³/512 + … , |x| < 4; original real domain x ≤ 4.
Look for these in your work
- Use the square root of the extracted constant.
- Keep the series interval distinct from the original domain.
Expand (9 + 6x)⁻¹ᐟ² through x³ and state the interval.
Give me a hint
Extract 9, whose power −1/2 is 1/3, and use u = 2x/3.
Compare my reasoning
- Rewrite as (1/3)(1 + 2x/3)⁻¹ᐟ².
- Using coefficients 1, −1/2, 3/8 and −5/16 gives 1/3 − x/9 + x²/18 − 5x³/162 + … .
- The condition |2x/3| < 1 gives |x| < 3/2; the outside factor applies to every term.
1/3 − x/9 + x²/18 − 5x³/162 + … , |x| < 3/2.
Look for these in your work
- Combine the fractional exponent with the constant and variable scales.
- Keep exact cubic coefficients and the strict transformed interval.
Common mistakes
- Raise the extracted constant to the exponent, including a negative or fractional exponent.
- Substitute the scaled variable into every power.
- The strict expansion interval and the original real function domain are different conditions.
What can you explain now?
Expand (4 + 2x)⁻¹ through x² and state its interval.
Compare with the explanation
1/4 − x/8 + x²/16 + … , |x| < 2.
Factor 4, use outside factor 1/4 and variable u = x/2; the strict condition |x/2| < 1 gives |x| < 2.
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Make it stick
Choose the outside multiplier and small variable before expanding. Explain why a real function can exist at an input where the local binomial expansion is not valid.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.