When you want the remainder of a polynomial P(x) divided by (x−k), you don't actually have to carry out the division at all. Just plug k in for x in P(x) and compute P(k) — that single number is the remainder. This is called the remainder theorem. It's a little startling that you can skip the whole long-division process and get the answer from one substitution.
Why does this work? Dividing P(x) by (x−k) can always be written as P(x) = (x−k)Q(x) + R, where Q(x) is the quotient and R (a constant) is the remainder. Substitute x=k into that equation, and (k−k)Q(k) becomes 0 and vanishes entirely, leaving just P(k) = R. That single substitution is a complete proof that the remainder equals P(k).
The quick way to actually carry out the remainder theorem is synthetic division. Pull out just the polynomial's coefficients, write them in a row, and repeat the process of multiplying by the divisor value k and adding — the coefficients of the quotient and the remainder come out in order. For example, dividing P(x)=x³−2x²−5x+6 by (x−1): write down the coefficients 1, −2, −5, 6 and work across, multiplying by k=1 each step.
The value left in the last box, 0, is the remainder. A remainder of 0 means the division comes out even by (x−1), which also means (x−1) is a factor of P(x) (the factor theorem). The two numbers before it, 1, −1, −6, become the coefficients of the quotient, so P(x) factors as P(x) = (x−1)(x²−x−6). If factoring in middle school stopped at quadratics, the remainder theorem and synthetic division open the door to factoring more complicated polynomials — cubics, quartics, and beyond.
Working with polynomials, you'll eventually run into multiplication formulas for higher powers too, beyond (a+b)² — (a+b)³, (a+b)⁴, and so on — and there's an interesting fact hiding here. The coefficient in front of each term when you expand (a+b)ⁿ turns out to be exactly the numbers in row n of Pascal's Triangle. In (a+b)³ = a³ + 3a²b + 3ab² + b³, you can check that the coefficients 1, 3, 3, 1 are exactly row three of Pascal's Triangle. Seeing polynomial multiplication formulas and number patterns connect like this is one of the fun parts of math.
When studying this with kids, it helps to introduce synthetic division as "a fast division that only needs the coefficients," then have them check it against an actual long division of the polynomial to confirm the quotient and remainder come out the same. On our activity page, you can compare the coefficients of (a+b)ⁿ side by side with Pascal's Triangle, then use sliders to change a polynomial's coefficients and the divisor value k and watch the synthetic division process line up with the remainder theorem in real time.