The Remainder Theorem

Pre-Calculus · Axiom Academy

Divide a polynomial by (x-c) and the leftover remainder is exactly P(c) — two different jobs, one answer. 1. Two Roads to the Same Number Divide any polynomial P(x) by (x-c) . Like dividing integers, you get a quotient q(x) and a remainder r — but because the divisor (x-c) has degree 1 , the remainder is just a constant. Substitute x=c into the division identity and the whole quotient term vanishes, leaving r alone equal to P(c) . Set x=c : the (x-c) factor zeroes out …so the remainder IS the value P(c) 2. Synthetic Division, Step by Step Let's verify the theorem with P(x) = 2x^3 - x^2 - 4 and c = 3 . Synthetic division turns the whole thing into a tidy march: bring down the leading coefficient, multiply by c , add down the next column, and repeat. The final number in the bottom row is the remainder. Bring down 2 ; multiply by 3 and add down each column. The last entry is 41 . P(x) = 2x^3 - x^2 - 4 has no x^1 term, so we write its coefficient as 0 . Every power needs a slot — that placeholder is what keeps the columns lined up. 3. The Remainder Is the Height at x=c Because the remainder equals P(c) , it's literally the height of the graph at x=c . That makes synthetic division a fast evaluator: instead of cubing a big number, you do a few multiply-and-adds. Watch the sweep land on x=5 for P(x) = x^3 - 4x^2 + 7x - 5 — the height it touches is the remainder.

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