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Polynomials

Degree > 2: factor/remainder theorem, Vieta for higher degree, polynomial division, roots

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PolynomialsAMC 10 Step by Step

Name: Date: October 28, 2026

11 problems, easiest first

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

Ask whether you need the roots at all. If swapping two roots doesn't change the question, use Vieta's Formulas directly.
For remainders, write $P=QD+R$ and plug in the roots of $D$ , which make $QD$ zero.
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 $P(4)=6+5=11$ .
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)

1.2025 AMC 10A #7
Suppose $a$ and $b$ are real numbers. When the polynomial $x^3+x^2+ax+b$ is divided by $x-1$ , the remainder is $4$ . When the polynomial is divided by $x-2$ , the remainder is $6$ . What is $b-a$ ?
1. A) $14$
2. B) $15$
3. C) $16$
4. D) $17$
5. E) $18$
2.2025 AMC 10B #10
Let $f(n)=n^3-5n^2+2n+8$ and $g(n)=n^3-6n^2+5n+12.$ What is the sum of all integers $n$ such that $\tfrac{f(n)}{g(n)}$ is an integer?
1. A) $2$
2. B) $3$
3. C) $4$
4. D) $5$
5. E) $6$
3.2023 AMC 10B #12
When the roots of the polynomial
$P(x) = (x-1)^1 (x-2)^2 (x-3)^3 \cdot \cdot \cdot (x-10)^{10}$
are removed from the number line, what remains is the union of $11$ disjoint open intervals. On how many of these intervals is $P(x)$ positive?
1. A) $3$
2. B) $7$
3. C) $6$
4. D) $4$
5. E) $5$
4.2023 AMC 10A #21
Let $P(x)$ be the unique polynomial of minimal degree with the following properties:
- $P(x)$ has a leading coefficient $1$ ,
- $1$ is a root of $P(x)-1$ ,
- $2$ is a root of $P(x-2)$ ,
- $3$ is a root of $P(3x)$ , and
- $4$ is a root of $4P(x)$ .
The roots of $P(x)$ are integers, with one exception. The root that is not an integer can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime integers. What is $m+n$ ?
1. A) $41$
2. B) $43$
3. C) $45$
4. D) $47$
5. E) $49$
5.2022 AMC 10A #16
The roots of the polynomial $10x^3 - 39x^2 + 29x - 6$ are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by $2$ units. What is the volume of the new box?
1. A) $\frac{24}{5}$
2. B) $\frac{42}{5}$
3. C) $\frac{81}{5}$
4. D) $30$
5. E) $48$
6.2021 AMC 10A #14
All the roots of the polynomial $z^6-10z^5+Az^4+Bz^3+Cz^2+Dz+16$ are positive integers, possibly repeated. What is the value of $B$ ?
1. A) ${-}88$
2. B) ${-}80$
3. C) ${-}64$
4. D) ${-}41$
5. E) ${-}40$
7.2010 AMC 10A #21
The polynomial $x^3 -ax^2 + bx -2010$ has three positive integer roots. What is the smallest possible value of $a$ ?
1. A) $78$
2. B) $88$
3. C) $98$
4. D) $108$
5. E) $118$
8.2022 AMC 10B #21
Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x + 2$ , and when $P(x)$ is divided by the polynomial $x^2 + 1$ , the remainder is $2x + 1$ . There is a unique polynomial of least degree with these two properties. What is the sum of the squares of the coefficients of that polynomial?
1. A) $10$
2. B) $13$
3. C) $19$
4. D) $20$
5. E) $23$
9.2017 AMC 10A #24
For certain real numbers $a$ , $b$ , and $c$ , the polynomial $g(x) = x^3 + ax^2 + x + 10$ has three distinct roots, and each root of $g(x)$ is also a root of the polynomial $f(x) = x^4 + x^3 + bx^2 + 100x + c.$ What is $f(1)$ ?
1. A) $-9009$
2. B) $-8008$
3. C) $-7007$
4. D) $-6006$
5. E) $-5005$
10.2019 AMC 10A #24
Let $p$ , $q$ , and $r$ be the distinct roots of the polynomial $x^3 - 22x^2 + 80x - 67$ . It is given that there exist real numbers $A$ , $B$ , and $C$ such that $\dfrac{1}{s^3 - 22s^2 + 80s - 67} = \dfrac{A}{s-p} + \dfrac{B}{s-q} + \frac{C}{s-r}$ for all $s\not\in\{p,q,r\}$ . What is $\tfrac1A+\tfrac1B+\tfrac1C$ ?
1. A) $243$
2. B) $244$
3. C) $245$
4. D) $246$
5. E) $247$
11.2010 AMC 10B #25
Let $a > 0$ , and let $P(x)$ be a polynomial with integer coefficients such that
$P(1) = P(3) = P(5) = P(7) = a$ , and
$P(2) = P(4) = P(6) = P(8) = -a$ .
What is the smallest possible value of $a$ ?
1. A) $105$
2. B) $315$
3. C) $945$
4. D) $7!$
5. E) $8!$

Polynomials — answersAMC 10 Step by Step

Name: Date: October 28, 2026

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Answers

#	Answer	The idea it turns on
1	E	Dividing by x - 1 leaves the value at x = 1 as the remainder; same for x - 2. That gives two easy equations in a and b.
2	A	Subtracting gives f − g = (n−4)(n+1), a factor of both. The fraction becomes (n−2)/(n−3), but n = 4 is out because g(4) = 0.
3	C	To the right of 10 every factor is positive. Moving left, P changes sign only when you pass a root with an odd power: 9, 7, 5, 3, 1.
4	D	Each clue is a value of P: P(1) is 1, and P is 0 at 0, 4 and 9. Those roots alone make P(1) equal 24, so add another root.
5	D	The new volume is (r + 2)(s + 2)(t + 2), where r, s, t are the roots. Plugging x = -2 into the polynomial gives this product almost directly.
6	A	The roots add to 10 and multiply to 16. Positive whole numbers with product 16 are powers of 2; only 2, 2, 2, 2, 1, 1 adds to 10.
7	A	The roots multiply to 2010 and add to a. One root must contain the prime 67; split the leftover 30 as evenly as possible, 5 and 6.
8	E	For the remainder on dividing by x² + 1, replace x² by −1. For x² + x + 1, replace x² by −x − 1 and x³ by 1.
9	C	Since f has all three roots of g, f equals g times (x - r). Matching x terms gives r = -90, so f(1) = g(1) times 91.
10	B	Clear the fractions and plug in each root: 1/A is the product of p minus each other root. Adding the three needs only the sum and pair-products of the roots.
11	B	P(x) - a is zero at 1, 3, 5, 7, so it has the factor (x-1)(x-3)(x-5)(x-7). Plugging in 2, 4, 6, 8 forces a to be a multiple of 315.

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