Polynomials

What you need to know

• Remainder theorem: dividing $P(x)$ by $x-a$ leaves remainder $P(a)$; factor theorem: $x-a$ divides $P(x)$ iff $P(a)=0$. Division by a degree-$d$ polynomial leaves a remainder of degree less than $d$.
• Vieta for a monic cubic $x^3+ax^2+bx+c$ with roots $r,s,t$: $r+s+t=-a$, $rs+st+tr=b$, $rst=-c$. Signs alternate, and every coefficient is divided by the leading coefficient.
• $P(1)$ is the sum of the coefficients and $P(0)$ the constant term.
• Rational root theorem: a rational root $\frac pq$ in lowest terms of an integer polynomial has $p\mid a_0$ and $q\mid a_n$.
• A degree-$n$ polynomial has $n$ roots with multiplicity; real coefficients force non-real roots into conjugate pairs.

How AMC 10 tests it

• Problems 8–15: the remainder when $x^{100}$ is divided by $x-1$, $x+1$, or $x^2-1$; write $P=QD+R$ and plug in the roots of $D$.
• Problems 12–18: a cubic with integer or given roots, asking for a coefficient or the sum of squares of the roots.
• "$P(1)=P(2)=P(3)=k$": build $P(x)-k$, which has known roots, then evaluate elsewhere.
• Problems 18–23: integer-coefficient polynomials using $a-b\mid P(a)-P(b)$.

Standard approaches

1. Ask whether you need the roots at all; if the question is symmetric in them, use Vieta directly.
2. For remainders, write $P=QD+R$ with $\deg R<\deg D$ and substitute the roots of $D$.
3. When $P$ is described by its values, build an auxiliary polynomial with known zeros, such as $P(x)-c$.
4. To find integer roots, test divisors of the constant term and reduce the degree by synthetic division.

Worked example

The polynomial $x^3+ax^2+bx+6$ has three integer roots (not necessarily distinct), one of which is $-3$. What is the sum of all possible values of $b$?

(A) $-7$ (B) $2$ (C) $4$ (D) $11$ (E) $15$

Let the other roots be $r$ and $s$. Vieta gives $(-3)rs=-6$, so $rs=2$: either $\{r,s\}=\{1,2\}$ or $\{-1,-2\}$. Vieta also gives
$$b=rs+(-3)(r+s)=2-3(r+s).$$
If $r+s=3$, then $b=-7$; if $r+s=-3$, then $b=11$. Both occur ($x^3-7x+6$ and $x^3+6x^2+11x+6$ factor as required). The sum is $-7+11=4$, so the answer is $\boxed{\textbf{(C)}\ 4}$.

Pitfalls

• Sign slips in cubic Vieta: the product of the roots of a monic cubic is $-c$, not $c$.
• Forgetting to divide by a leading coefficient other than $1$.
• Assuming the remainder on division by a quadratic is a constant; it can be linear.
• Ignoring repeated or non-real roots when counting.

Traps that recur

• Trying to find the roots or the individual constants A, B, C numerically instead of expressing 1/A + 1/B + 1/C symmetrically and using Vieta. (2019 AMC 10A #24)
• Stopping at the lower bound and forgetting to confirm that an integer polynomial Q actually exists for a = 315, or using the product 15915*105 instead of the lcm. (2010 AMC 10B #25)
• Trying a quadratic P first and not recognizing the contradiction, or expanding P and forgetting the constant term 3 when summing squares. (2022 AMC 10B #21)
• Trying to find the roots of g, or solving for b and c by full expansion and making an arithmetic slip; f(1) = g(1)(1 - r) needs only a and r. (2017 AMC 10A #24)