The coefficients of (a+b)ⁿ are exactly Pascal's Triangle
See why the coefficients you get from expanding (a+b)ⁿ match Pascal's Triangle exactly, then use synthetic division to verify the remainder theorem — that dividing a polynomial by (x−k) always leaves a remainder equal to P(k).
Expanding (a+b)ⁿ produces several terms, and the coefficient in front of each one happens to match, exactly, the numbers in row n of Pascal's Triangle. In (a+b)³ = a³+3a²b+3ab²+b³, the coefficients 1, 3, 3, 1 are exactly row three of the triangle.
The remainder when a polynomial P(x) is divided by (x−k) comes out immediately if you just substitute P(k) — no need to carry out the division at all (the remainder theorem). Synthetic division is a way to run that same division quickly, using only the coefficients.
Use the slider to change n in (a+b)ⁿ and compare the coefficients side by side with Pascal's Triangle, then change the polynomial and the value of k to watch the synthetic division process step by step.
Binomial coefficients — expanding (a+b)ⁿ gives terms aⁿ, aⁿ⁻¹b, …, bⁿ, and the coefficient in front of each one is exactly the numbers in row n of Pascal's Triangle.
Power n3
a2
b1
Remainder theorem — the remainder when polynomial P(x) is divided by (x−k) comes out immediately when you compute P(k), with no need to carry out the division. Synthetic division runs that same division quickly using only the coefficients. If the remainder is 0, that means (x−k) is a factor of P(x) (the factor theorem).