To check an expansion, substitute a simple value of x into both the original bracket and your answer. The two must give the same number. The easiest choices are x = 1, x = −1 and x = 0.
This lesson builds on tracking signs and works with any expansion from expanding a power systematically.
What does each substitution test?
- x = 1 turns every power of x into 1, so you compare the bracket’s value with the sum of all your coefficients.
- x = −1 flips the sign of odd powers. This checks whether your signs alternate correctly.
- x = 0 leaves only the constant term, so you compare it with the bracket evaluated at 0.
Each check is short arithmetic you can do in your head or in a margin.
A routine that takes thirty seconds
- Pick x = 1 first, and x = −1 if the bracket contains a minus sign.
- Evaluate the bracket at that value.
- Evaluate your expansion at the same value, adding the signed terms.
- Compare. If they differ, find the error before moving on.
- Repeat with a second value when the answer matters a lot.
Worked example
A student expands (3x − 1)⁴ and writes 81x⁴ − 108x³ + 54x² − 12x + 1. Check it.
Check at x = 1: the bracket gives (3 − 1)⁴ = 2⁴ = 16. The expansion gives 81 − 108 + 54 − 12 + 1 = 16. They agree.
Check at x = −1: the bracket gives (−3 − 1)⁴ = (−4)⁴ = 256. The expansion gives 81 + 108 + 54 + 12 + 1 = 256. They agree again.
Check at x = 0: the bracket gives (−1)⁴ = 1, and the constant term is 1.
Three checks pass, so the expansion can be trusted.
The mistake to watch for
The check is most useful when something is wrong.
Mistaken expansion: 81x⁴ − 12x³ + 54x² − 12x + 1
Here the student wrote the x³ coefficient as −12 instead of −108. At x = 1 the expansion gives 81 − 12 + 54 − 12 + 1 = 112, but the bracket gives 16.
A mismatch tells you an error exists but not where. Recheck each term: here 4 × (3x)³ × (−1) = 4 × 27x³ × (−1) = −108x³. The student had used 3x³ in place of (3x)³, the same slip seen in the first lesson.
Check yourself
Try these, then open each answer.
1. Without expanding, what is the sum of the coefficients in (2x + 1)⁵?
Show answer
Put x = 1: (2 + 1)⁵ = 3⁵ = 243.
2. Without expanding (x − 2)⁷, find the sum of its coefficients, and the value it takes at x = −1.
Show answer
At x = 1: (1 − 2)⁷ = (−1)⁷ = −1. At x = −1: (−1 − 2)⁷ = (−3)⁷ = −2187.
3. A student claims (x + 3)⁴ = x⁴ + 12x³ + 36x² + 108x + 81. Use x = 1 to test it.
Show answer
The bracket gives 4⁴ = 256. The claim gives 1 + 12 + 36 + 108 + 81 = 238. These differ, so the expansion is wrong. The x² coefficient should be 6 × 9 = 54, not 36. The correct expansion is x⁴ + 12x³ + 54x² + 108x + 81, which sums to 256.
Where this leads next
Try the full binomial expansion practice set, using a check after every expansion. Keeping a record in the mistake log and retest queue shows which kind of slip the check most often catches.
Learning to trust your own checks is one of the quickest ways to gain confidence. We build that habit with students in online one-to-one Additional Mathematics tuition.