Polynomials

What you need to know

• Remainder Theorem. Dividing a polynomial $P(x)$ by $x-a$ leaves the remainder $P(a)$. Example: $x^2+3x+1$ divided by $x-2$ leaves $4+6+1=11$.
• Factor Theorem. $x-a$ is a factor of $P(x)$ exactly when $P(a)=0$. Example: for $P(x)=x^3-6x^2+11x-6$, $P(1)=0$, so $x-1$ is a factor.
• The remainder has a lower degree than what you divide by. So dividing by $x-a$ leaves a number, and dividing by a quadratic leaves a remainder like $px+q$. Example: write $x^{100}=Q(x)(x^2-1)+px+q$, where $Q(x)$ is the quotient. Plugging in $x=1$ and $x=-1$ gives $p+q=1$ and $-p+q=1$, so the remainder is $1$.
• Vieta's Formulas for a cubic. If $x^3+ax^2+bx+c$ has roots $r$, $s$, $t$, then $r+s+t=-a$, $rs+st+tr=b$, and $rst=-c$. The signs go minus, plus, minus. Example: $x^3-6x^2+11x-6$ has roots $1,2,3$, and $1+2+3=6$, $2+6+3=11$, $1\cdot2\cdot3=6$.
• If the leading coefficient is not $1$, divide every coefficient by it first. Example: the roots of $2x^3-6x^2+4x-10$ add to $\frac62=3$ and multiply to $\frac{10}2=5$.
• $P(1)$ is the sum of the coefficients, and $P(0)$ is the constant term. Example: for $P(x)=3x^4-2x^2+5x+7$, $P(1)=3-2+5+7=13$ and $P(0)=7$.
• Rational Root Theorem. If a polynomial has integer coefficients and a fraction $\frac pq$ in lowest terms is a root, then $p$ divides the constant term and $q$ divides the leading coefficient. Example: $2x^3+x^2-7x-6$ has roots $2$, $-1$, $-\frac32$; each top divides $6$ and each bottom divides $2$.
• Counting roots. A polynomial of degree $n$ has exactly $n$ roots, counting each repeated root as often as it repeats (its multiplicity) and including non-real roots. With real coefficients, non-real roots come in pairs $p+qi$ and $p-qi$ (conjugates), where $i=\sqrt{-1}$. Example: $(x-2)^2(x+1)$ has roots $2,2,-1$, and $(x-1)(x^2+1)$ has roots $1,i,-i$.

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 $P=QD+R$ (quotient $Q$, divisor $D$, remainder $R$) 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 the squares of the roots.
• "$P(1)=P(2)=P(3)=k$": build $P(x)-k$, which has the known roots $1,2,3$, then evaluate it somewhere else.
• Problems 18–23: polynomials with integer coefficients, using the fact that $a-b$ divides $P(a)-P(b)$ for different integers $a$ and $b$.

Standard approaches

1. Ask whether you need the roots at all. If swapping two roots doesn't change the question, use Vieta's Formulas directly.
2. For remainders, write $P=QD+R$ and plug in the roots of $D$, which make $QD$ zero.
3. When $P$ is described by its values, build a helper polynomial with known roots, such as $P(x)-c$. Example: if $P$ is a cubic with leading coefficient $1$ and $P(1)=P(2)=P(3)=5$, then $P(x)-5=(x-1)(x-2)(x-3)$, so $P(4)=6+5=11$.
4. To find integer roots, test the divisors of the constant term, positive and negative. Divide out each root found (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) $15$

Call the other two roots $r$ and $s$. By Vieta's Formulas the roots multiply to $-6$, so $(-3)rs=-6$ and $rs=2$. The integer pairs with product $2$ are $1,2$ and $-1,-2$.

Vieta's Formulas also say $b$ is the sum of the products of pairs of roots:
$$b=rs+(-3)r+(-3)s=2-3(r+s).$$
If the roots are $1,2$, then $r+s=3$ and $b=2-9=-7$. If they are $-1,-2$, then $r+s=-3$ and $b=2+9=11$.

Both cases happen: $(x+3)(x-1)(x-2)=x^3-7x+6$ and $(x+3)(x+1)(x+2)=x^3+6x^2+11x+6$.

The sum is $-7+11=4$. The answer is $\boxed{\textbf{(C)}\ 4}$.

Pitfalls

• Sign slips in Vieta's Formulas for a cubic: the roots of $x^3+ax^2+bx+c$ multiply to $-c$, not $c$.
• Forgetting to divide by a leading coefficient that is not $1$: the roots of $2x^3-6x^2+4x-10$ add to $3$, not $6$.
• Assuming the remainder on division by a quadratic is a number. It can be $px+q$: $x^3=x(x^2-1)+x$ leaves remainder $x$.
• Forgetting repeated or non-real roots when counting: $(x-2)^2(x+1)$ has three roots but only two different ones.

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)