A binomial expansion becomes reliable when three patterns stay aligned: the numerical coefficients, the powers of the two terms and the signs. Writing these separately before simplifying is usually safer than trying to remember a finished polynomial. It also makes a wrong term easier to locate.
This lesson covers positive-integer powers such as (3 − 2x)5. Each complete expansion has finitely many terms. We also use part of a finite expansion to estimate a number, stating which terms were kept and how far the estimate is from the exact value. Negative and fractional indices, and infinite series, are outside this lesson.
Build coefficients with Pascal’s triangle
Start each row with 1 and end it with 1. Every interior entry is the sum of the two entries immediately above it. Label the top row n = 0 so that row n = 5 belongs to a fifth power. Counting the top row as “row one” without an exponent label is a common source of an off-by-one error.
| Exponent n | Coefficients in order |
|---|---|
| 0 | 1 |
| 1 | 1, 1 |
| 2 | 1, 2, 1 |
| 3 | 1, 3, 3, 1 |
| 4 | 1, 4, 6, 4, 1 |
| 5 | 1, 5, 10, 10, 5, 1 |
For (a + b)5, attach these six coefficients to a5, a4b, a3b2, a2b3, ab4 and b5. The exponent on a falls by one at each step; the exponent on b rises by one. Before substitution, the two exponents in every term add to five.
The general term is Tr+1 = C(n,r)an−rbr, for r from 0 through n, where C(n,r) = n!/[r!(n − r)!]. Here n! means the product of positive integers through n, and 0! = 1. The first term uses r = 0, so the fourth term uses r = 3.
Worked case 1: expand (3 − 2x)5
Use a = 3 and b = −2x. Keep the negative sign inside the bracket being raised to a power. The fifth-power coefficient row is 1, 5, 10, 10, 5, 1. The complete assembly is shown below; the powered second term supplies each sign.
| r | Unsimplified term | Simplified term |
|---|---|---|
| 0 | 1 × 35 | 243 |
| 1 | 5 × 34 × (−2x) | −810x |
| 2 | 10 × 33 × (−2x)2 | 1080x2 |
| 3 | 10 × 32 × (−2x)3 | −720x3 |
| 4 | 5 × 3 × (−2x)4 | 240x4 |
| 5 | (−2x)5 | −32x5 |
Therefore (3 − 2x)5 = 243 − 810x + 1080x2 − 720x3 + 240x4 − 32x5. Notice that the fourth term is −720x3, while its coefficient is −720. The phrase “coefficient of x cubed” asks for the number, including its sign.
At x = 1, the original expression is 15 = 1. The polynomial gives 243 − 810 + 1080 − 720 + 240 − 32 = 1. At x = −1, the original expression is 55 = 3125; the polynomial gives 243 + 810 + 1080 + 720 + 240 + 32 = 3125. These substitutions catch many errors, although two matching values alone do not prove a polynomial identity.
Worked case 2: estimate a number using three terms
Estimate 1.025 using the first three terms of (1 + x)5, in ascending powers of x. The full expansion is 1 + 5x + 10x2 + 10x3 + 5x4 + x5. Here x = 0.02; do not substitute 1.02 for x, because the expression already contains the leading 1.
The retained terms give 1 + 5(0.02) + 10(0.02)2 = 1 + 0.10 + 0.004 = 1.104. This is an estimate, so write 1.025 ≈ 1.104. The omitted terms are 0.00008, 0.0000008 and 0.0000000032. Their sum is 0.0000808032.
Thus the exact finite-expansion value is 1.1040808032. The estimate underestimates it by 0.0000808032; both values round to 1.104 to three decimal places. Agreement after rounding is a checked result here, not a rule that three terms always guarantee three correct decimal places.
Small powers of 0.02 shrink quickly, which explains why truncation works well in this example. Before trusting another estimate, inspect its omitted terms. Their size and signs determine the error. A calculator result is useful as an independent comparison, but it should not replace the requested expansion method.
Worked case 3: recover a missing coefficient
In the expansion of (2 + kx)4, the coefficient of x is −96. Find k, then write the coefficient of x2. The x term is 4 × 23 × kx = 32kx. Hence 32k = −96 and k = −3.
The x2 term is 6 × 22 × (kx)2 = 24k2x2. With k = −3, its coefficient is 24 × 9 = 216. It is positive because the entire negative value is squared.
For a full check, (2 − 3x)4 = 16 − 96x + 216x2 − 216x3 + 81x4. At x = 1 the sum is 1, matching (−1)4. At x = −1 the sum is 625, matching 54.
Diagnose errors before expanding again
Pascal’s coefficients are only one factor in a term. For example, 10 is the Pascal coefficient in the x2 term of the fifth-power example, but the coefficient of x2 after substitution is 1080. The powers of 3 and −2 still have to be included.
Also distinguish total degree in a and b from degree in x after substitution. The generic term a3b2 has total degree five, but if a = 3 and b = −2x, it becomes a multiple of x2. There is no missing x3: a was a constant. For a subtraction, alternating signs follow from even and odd powers of b; they should not be added as an unrelated decoration afterward.
Three practice tasks with full solutions
Practice 1: expand (2 + x)4
Use the row 1, 4, 6, 4, 1. The terms are 24, 4 × 23x, 6 × 22x2, 4 × 2x3 and x4. Therefore the expansion is 16 + 32x + 24x2 + 8x3 + x4. At x = 1 it totals 81, equal to 34.
Practice 2: expand (1 − 3x)3
The cubic row gives 1 + 3(−3x) + 3(−3x)2 + (−3x)3 = 1 − 9x + 27x2 − 27x3. At x = 1 this is −8, matching (−2)3. The square term stays positive.
Practice 3: expand and estimate 0.984
Expand (1 − x)4 = 1 − 4x + 6x2 − 4x3 + x4. At x = 0.02, the first three terms give 0.9224. The omitted total is −0.000032 + 0.00000016 = −0.00003184, so the exact value is 0.92236816. The estimate is too high by 0.00003184; both round to 0.9224 to four decimal places.
Frequently asked questions
How many terms should a complete expansion have?
The generic expansion of a positive-integer power n has n + 1 terms, before any special simplification or zero coefficients.
Why does the first term use r = 0?
It contains b to power zero. The term number is one more than r, so T₁ corresponds to r = 0.
Are Pascal’s coefficients sometimes negative?
The entries shown here are positive. Negative signs come from powers of a negative term substituted into the expansion.
Is a coefficient the whole term?
No. In −720x³, the coefficient of x³ is −720; the whole term includes x³.
Does a three-term estimate equal the full expression?
Generally no. Omitted terms create an error. State the retained terms and use an approximation sign.
Do two substitutions prove the expansion?
No. They are useful checks. Correct term construction or complete multiplication establishes the identity.
Can this finite pattern be used unchanged for fractional powers?
No. This lesson assumes positive integers. Other indices need a different treatment and conditions beyond this scope.
Extend your practice
Choose a binomial, build its coefficient row, then check the finished expression at two values. For further navigation, use the Mathematics topical revision collection. Some resources may require membership; a specific expansion file is not promised.
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