Method of Undetermined Coefficients

Given information about a polynomial's zeros and/or its value at specific points, we can construct it by writing it with unknown ('undetermined') coefficients and using the equality test of §2.6 -- match same-power coefficients on both sides of an equation -- to pin those coefficients down.

Two equivalent approaches. To build a quadratic $f(x)=ax^2+bx+c$ satisfying $f(0)=1,\ f(-2)=0,\ f(1)=0$: either substitute all three conditions directly and solve the resulting linear system for $a,b,c$; or, since $x=-2,1$ are already known zeros, write $f(x)=d(x+2)(x-1)$ for an unknown constant $d$ and use the remaining condition $f(0)=1$ to solve for $d$ -- the zero-factor form is usually the faster route whenever the zeros are already given.

Constructing from a mix of real and irrational zeros. If an irrational zero like $1+\sqrt3$ is given for a polynomial with rational coefficients, its conjugate $1-\sqrt3$ must also be a zero (§2.6.1's Note); build the factor $[(x-1)^2-3]$ from the conjugate pair before bringing in any other given zero and an extra condition (like a specified function value) to fix the overall scale constant.

Using the coefficient-matching idea for a divisibility identity. E.g. to prove $ap+q=0$ given $f(x)=x^3-3px+2q$ is divisible by $g(x)=x^2+2ax+a^2$: since $\deg f=3,\deg g=2$, the quotient must be linear, $f(x)=(x+b)g(x)$; expand and match same-power coefficients on both sides to solve for $b,p,q$ in terms of $a$, then combine to get the required identity.

Undetermined coefficients also finds closed forms for sums. To find $S(n)=1+2+\cdots+n$: first bound it above by $n^2$ crudely; then, since the growth pattern looks quadratic, POSIT $S(n)=a+bn+cn^2$ for unknown constants, use the fact that $S(n+1)-S(n)=n+1$ to match coefficients and solve for $b,c$, and use $S(1)=1$ to solve for $a$ -- giving the closed form $S(n)=\dfrac{n(n+1)}2$.

Roots and multiplicity read off from a factored form. For $f(x)=(x-1)^3(x+1)^2(x+5)=0$, the roots are $1$ (multiplicity 3), $-1$ (multiplicity 2), and $-5$ (multiplicity 1, a simple root) -- multiplicities are visible directly from the exponents once $f$ is fully factored, without any further work. …