Q.Using synthetic division, find the quotient and remainder when $f(x)=2x^4-3x^3+4x^2-5x+6$ is divided by $x-1$.

For a polynomial $f(x)=a_0x^n+a_1x^{n-1}+\cdots+a_n$ ($a_0\neq0$) and any number $a$, dividing $f(x)$ by the linear polynomial $x-a$ always leaves a constant remainder, and that constant is exactly $f(a)$. This is the Remainder Theorem:

for a unique quotient $q(x)$ of degree $n-1$.

Why it is true. Write $f(x)-f(a) = a_0(x^n-a^n)+a_1(x^{n-1}-a^{n-1})+\cdots+a_{n-1}(x-a)$. Every bracket $x^k-a^k$ factors as $(x-a)(x^{k-1}+x^{k-2}a+\cdots+a^{k-1})$, so it is divisible by $x-a$; hence $f(x)-f(a)$ is divisible by $x-a$, and calling the quotient $q(x)$ gives $f(x)=(x-a)q(x)+f(a)$. Setting $x=a$ on both sides confirms the constant term is indeed $f(a)$.

Synthetic division is the fast bookkeeping method that carries out this division and reads off $q(x)$'s coefficients directly. If $f(x)=a_0x^n+a_1x^{n-1}+\cdots+a_n$, the quotient's coefficients $b_0,b_1,\ldots,b_{n-1}$ satisfy the recursion $b_0=a_0$ and $b_k=a_k+a\,b_{k-1}$ for $k=1,\ldots,n-1$, and the final remainder is $r=a_n+a\,b_{n-1}=f(a)$. Arranging this as 'bring down the first coefficient, multiply by $a$ and add to the next, repeat' is exactly synthetic division. In particular $f(a)=0$ exactly when $x-a$ divides $f(x)$ exactly -- this is how a trial root is confirmed before the equation is depressed to a lower degree.