A coefficient question asks for one part of a polynomial, not its value at a convenient input. Thinking about which factors supply x helps explain the binomial coefficient directly.
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Choose the factors supplying x
Consider (1 + 2x)5 as five multiplied factors. To obtain an x3 term, choose 2x from exactly three factors and 1 from the remaining two.
There are C(5,3) = 10 ways to choose those three factors. Each choice contributes (2x)3 = 8x3, so the total term is 80x3 and the coefficient is 80.
Separate three different quantities
The requested coefficient is 80. The corresponding term is 80x3. The value of the whole polynomial at x = 1 is 35 = 243, which includes all coefficients.
Substituting x = 1 is a useful overall expansion check, but it cannot by itself isolate the coefficient of x3.
Adapt the reasoning to a sign change
For (1 - 2x)5, the same choice count applies, but each selection contributes (-2)3x3. The coefficient is therefore -80.
Record the exponent, the number of choices and the numerical contribution separately. This prevents losing a power of 2 or an odd negative sign. The example is original teaching material for AMC 12 preparation; current MAA information should be used for the competition itself.
Continue reading: AMC 12 Trigonometry: Solve Equations Without Losing Solutions.
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Official information and further reading: Source 1
