Synthetic Division and Solving Equations Using Extra Root Conditions

Dividing a polynomial by $(x-a)$ using the Remainder Theorem's coefficient pattern is called synthetic division — a compact bookkeeping scheme that avoids writing out the full long division. Starting from $a_0x^n+a_1x^{n-1}+\cdots+a_n$ divided by $(x-a)$, you generate the quotient's coefficients $b_0,b_1,\dots,b_{n-1}$ by the rule $b_0=a_0$ and $b_k = a_k + a\,b_{k-1}$, with the final remainder $R = a_n + a\,b_{n-1}$. The same idea extends to dividing by a quadratic $x^2-px-q$ in two passes, which is handy when you know a pair of roots satisfies a known quadratic relation rather than a single linear one.

Synthetic division becomes the workhorse for a very practical class of problem: you are given a relation the roots must satisfy (one root is a known number, two roots are equal, the roots are in arithmetic or geometric progression, one root is a fixed multiple of another, etc.), and asked to find every root. The method is always the same three-step pattern: (1) translate the given relation, together with the standard sum/product relations from Section 4.2, into a small system of equations in the unknown root-parameters; (2) solve that system for the parameters; (3) use synthetic division to confirm and peel off the roots one at a time, reducing the degree until what remains is a quadratic or linear equation you can finish by inspection.

Worked example. Solve $x^3 - 7x^2 + 14x - 8 = 0$, given that its roots are in geometric progression.

Let the roots be $\frac{\alpha}{r}, \alpha, \alpha r$. Their product is $\alpha^3 = -(-8) = 8$ (using $s_3 = \alpha\beta\gamma = -p_3$), so $\alpha = 2$. Since $\alpha=2$ is claimed to be a root, synthetic division by $(x-2)$ should leave zero remainder: …