When a polynomial is divided by a quadratic, the remainder can be linear. Evaluating at roots of the divisor can determine that remainder without expanding a long division.
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Write the form of the remainder first
For the original polynomial P(x) = x7 + 2x4 + 3, find the remainder on division by x2 - 1. Write P(x) = (x2 - 1)Q(x) + ax + b.
The remainder has degree less than two, so ax + b is the general form. Beginning with only a constant would discard a possibility before using the data.
Use the two zeros of the divisor
At x = 1, the multiple of the divisor vanishes. Since P(1) = 6, we have a + b = 6. At x = -1, P(-1) = 4, giving -a + b = 4.
Adding the equations gives b = 5, and then a = 1. The remainder is x + 5. A single evaluation would leave two unknown coefficients with only one condition.
Check by reducing powers
Modulo x2 - 1, we have x2 equivalent to 1. Therefore x7 reduces to x and x4 reduces to 1. The expression reduces to x + 2 + 3 = x + 5.
This is an algebraic congruence modulo a polynomial. It does not assert that x2 = 1 for every real value of x.
Notice when the root method needs more care
Two distinct roots provide two independent conditions for a linear remainder. A repeated-root divisor such as (x - 1)2 does not give two different conditions merely by substituting x = 1 twice; additional information is needed.
This original AMC 12 exercise is not an official contest problem. In a written solution, state the allowed remainder degree and why each substitution removes the unknown quotient before solving the coefficient equations.
Continue reading: AMC 12 Algebra: Find One Binomial Coefficient Without Expanding Everything.
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