Polynomials

1.Division, the remainder theorem and the factor theorem

Every polynomial $P(x)$ with real coefficients can be divided by a linear polynomial $x-a$. The division algorithm says $P(x)=(x-a)Q(x)+R$, where $Q$ is a polynomial of degree one less than $P$ and $R$ is a constant, because the remainder must have degree less than $1$. This single identity is the source of the two most useful facts about polynomials in contest problems.

Substitute $x=a$ into the identity. The term $(x-a)Q(x)$ vanishes, so $P(a)=R$. This is the remainder theorem: the remainder on division by $x-a$ equals the value $P(a)$. It means you can read a remainder without performing any long division at all. The factor theorem is the special case $R=0$: the linear factor $x-a$ divides $P(x)$ exactly if and only if $P(a)=0$, that is, if and only if $a$ is a root.

The same reasoning works for any divisor. If $D(x)$ has degree $d$, then $P(x)=D(x)Q(x)+R(x)$ with $\deg R<d$. For $D(x)=(x-a)(x-b)$ with $a\ne b$, the remainder is a line $R(x)=px+q$, and substituting $x=a$ and $x=b$ gives the two equations $P(a)=pa+q$ and $P(b)=pb+q$. Solving them is usually quicker than dividing. This is the first appearance of an idea we will return to in the interpolation concept: a low-degree polynomial is determined by enough values.

Two practical consequences matter on the test. First, if $P$ has integer coefficients and $a$ is an integer, then $P(a)$ is the remainder, so a divisibility condition such as $(x-a)\mid P(x)$ becomes the single equation $P(a)=0$. Second, a nonzero polynomial of degree $n$ has at most $n$ roots, which is proved by repeatedly applying the factor theorem: each root peels off one linear factor, and the degree drops by one each time. This counting argument is the reason that two polynomials of degree at most $n$ agreeing at $n+1$ points must be equal, a fact used heavily in interpolation.

• $P(x)=(x-a)Q(x)+P(a)$ for every polynomial $P$
• $(x-a)\mid P(x)$ if and only if $P(a)=0$
• Remainder on division by $(x-a)(x-b)$ is $px+q$ with $pa+q=P(a)$, $pb+q=P(b)$
• A nonzero polynomial of degree $n$ has at most $n$ roots

• Taking the remainder of a quadratic divisor to be a constant; it is a linear polynomial $px+q$ and needs two conditions.
• Substituting the root into $P$ and forgetting that the answer is the remainder, not the quotient.

2.Vieta's formulas for every degree

Vieta's formulas translate between the roots of a polynomial and its coefficients. Suppose $P(x)=x^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0$ is monic, and let its roots be $r_1,\dots,r_n$, counted with multiplicity. Then $P(x)=(x-r_1)(x-r_2)\cdots(x-r_n)$. Expanding the product, the coefficient of $x^{n-k}$ is $(-1)^k e_k$, where $e_k$ is the $k$-th elementary symmetric polynomial of the roots, that is, the sum of all products of $k$ distinct roots. So $e_1=r_1+\cdots+r_n$, $e_2=\sum_{i<j}r_ir_j$, and in general $e_n=r_1r_2\cdots r_n$.

The proof is just expansion. Choosing one of the $n$ factors to contribute $x$ and the others to contribute $-r_i$ produces the term $(-1)^k e_k x^{n-k}$, because each product of $k$ roots appears exactly once. You do not need to memorise the sign pattern: it alternates, and the constant term is $(-1)^n$ times the product of all roots.

For a quadratic $ax^2+bx+c$ with roots $r,s$, dividing by $a$ gives $r+s=-b/a$ and $rs=c/a$. For a cubic $x^3+px^2+qx+t$ with roots $u,v,w$, we get $u+v+w=-p$, $uv+vw+wu=q$, and $uvw=-t$. Notice the signs: the sum flips sign, the pairwise sum does not, and the product flips again. Students often write $uvw=t$, which is the most common sign error on this topic.

Vieta is valuable because most questions about roots are questions about symmetric functions of roots. Any expression symmetric in the roots can be written in terms of the $e_k$. For example $\sum r_i^2=e_1^2-2e_2$, since $(\sum r_i)^2$ counts each square once and each cross term $2r_ir_j$ once. Once you see that a quantity is symmetric, you can compute it from coefficients alone, which is the whole point of the formulas. When the leading coefficient is not $1$, divide the whole polynomial by it first.

• For monic $P$ of degree $n$: $\sum r_i=-c_{n-1}$, $\sum_{i<j} r_ir_j=c_{n-2}$, $\prod r_i=(-1)^n c_0$
• For $ax^n+\cdots$ divide by $a$ first: $e_1=-\frac{b}{a}$, $e_2=\frac{c}{a}$ in a quadratic $ax^2+bx+c$
• Expanding $\sum r_i^2=e_1^2-2e_2$ shows how symmetric sums reduce to coefficients

• Forgetting the sign: for a cubic, $uvw=-t$ and $u+v+w=-p$ when the polynomial is $x^3+px^2+qx+t$.
• Applying Vieta to a non-monic polynomial without dividing by the leading coefficient first.

3.Power sums and Newton's identities

A power sum is $p_k=r_1^k+r_2^k+\cdots+r_n^k$. Vieta gives $p_1=e_1$ directly, but higher power sums are not elementary symmetric functions, so they need a recursion. Newton's identities provide it. For $1\le k\le n$ they read $p_k-e_1p_{k-1}+e_2p_{k-2}-\cdots+(-1)^{k-1}e_{k-1}p_1+(-1)^k k e_k=0$, and for $k>n$ they simplify to $p_k=e_1p_{k-1}-e_2p_{k-2}+\cdots+(-1)^{n-1}e_np_{k-n}$.

The derivation is short. Each root $r_i$ satisfies $P(r_i)=0$, so $r_i^n=-(c_{n-1}r_i^{n-1}+\cdots+c_0)$. Multiply by $r_i^{k-n}$ for $k\ge n$ and sum over $i$. The left side gives $p_k$, and the right side gives a linear combination of earlier power sums with the coefficients of $P$. Since the coefficients are $\pm e_j$ by Vieta, the recursion has exactly the form above. For small $k$ the boundary terms with $ke_k$ come from the fact that $p_k$ is not yet determined by the recursion alone.

A quadratic is the cleanest example. With roots $r,s$, $e_1=r+s$ and $e_2=rs$, so $p_1=e_1$, $p_2=e_1^2-2e_2$, and for $k\ge 3$ the recursion is $p_k=e_1p_{k-1}-e_2p_{k-2}$. For $x^2-5x+3$ we have $p_1=5$, $p_2=25-6=19$, and $p_3=5\cdot 19-3\cdot 5=80$. You can confirm the last value by expanding $(r+s)^3-3rs(r+s)=125-45=80$.

Newton's identities are useful whenever a problem asks for $\sum r_i^k$ with $k$ larger than $2$. Computing the roots is usually impossible on an exam, but the recursion lets you build $p_k$ one step at a time from the coefficients. A common pattern is a problem with a polynomial of degree $3$ or $4$ and a request for $p_{10}$ or a combination such as $\sum r_i^{5}-\sum r_i^{4}$. Write the recursion once, tabulate $p_1,\dots,p_k$, and be careful with the signs, which alternate.

• $p_1=e_1$, $p_2=e_1^2-2e_2$, $p_3=e_1^3-3e_1e_2+3e_3$
• For $k>n$: $p_k=e_1p_{k-1}-e_2p_{k-2}+\cdots+(-1)^{n-1}e_np_{k-n}$
• For a quadratic with roots $r,s$: $r^k+s^k=e_1(r^{k-1}+s^{k-1})-e_2(r^{k-2}+s^{k-2})$

• Using $p_k=e_1p_{k-1}-e_2p_{k-2}$ for $k=2$, which needs the separate formula $p_2=e_1^2-2e_2$.
• Mixing up the sign of $e_2$ in the recursion: the minus sign is between terms, not on $e_2$ itself.

4.The rational root theorem

Suppose a polynomial $a_nx^n+\cdots+a_1x+a_0$ has integer coefficients, and suppose $x=p/q$ is a root written in lowest terms. Substitute and multiply through by $q^n$. Every term except $a_0q^n$ contains a factor of $p$, so $p\mid a_0q^n$. Since $\gcd(p,q)=1$, we get $p\mid a_0$. Symmetrically, multiplying the other way shows $q\mid a_n$. This is the rational root theorem: every rational root has numerator dividing the constant term and denominator dividing the leading coefficient.

The theorem does not find roots; it produces a finite list of candidates $\pm p/q$ with $p\mid a_0$ and $q\mid a_n$. You then test each candidate. For a monic polynomial the denominator must be $1$, so the rational roots are integers dividing $a_0$. That is why monic integer polynomials are so pleasant: the candidate list is just the divisors of the constant term, and a rational root is automatically an integer.

The theorem is most useful when combined with Vieta. Suppose you know one rational root $r$ from testing candidates. Then the factor theorem gives $P(x)=(x-r)Q(x)$, and Vieta applied to $Q$ reveals the remaining roots' sum and product. In the example below, testing candidates finds three rational roots, and their sum is read off from the coefficient of $x^2$ divided by the leading coefficient, without solving anything further.

Do not forget the converse direction, which is false. A polynomial may have irrational or complex roots, and the theorem says nothing about them. Also, a root being rational does not force it to be an integer unless the polynomial is monic. Students often test candidates $\pm p/q$ without restricting to divisors of $a_0$ and $a_n$, which gives a much longer and error-prone list. Write the candidates explicitly before testing, and cross them off using the factor theorem.

• If $p/q$ (lowest terms) is a root of $a_nx^n+\cdots+a_0$, then $p\mid a_0$ and $q\mid a_n$
• A rational root of a monic integer polynomial is an integer dividing the constant term
• After finding a rational root $r$, divide out $(x-r)$ and apply Vieta to the quotient

• Listing candidates $\pm p/q$ without requiring $p\mid a_0$ and $q\mid a_n$ first, which makes the list far too long.
• Assuming all roots are rational because one is, and forgetting the remaining quotient may have irrational roots.

5.Integer-coefficient tricks: $a-b$ divides $P(a)-P(b)$

Let $P$ have integer coefficients. For any integers $a$ and $b$, the difference $P(a)-P(b)$ is divisible by $a-b$. The reason is that each power satisfies $a^k-b^k=(a-b)(a^{k-1}+a^{k-2}b+\cdots+b^{k-1})$, and the second factor is an integer because $a$ and $b$ are integers. Multiplying by the integer coefficients and summing gives $P(a)-P(b)=(a-b)\cdot N$ for some integer $N$. This is the basic divisibility principle for integer polynomials.

The consequence is a strong restriction on values. If $P(3)=7$, then $P(8)-P(3)$ must be divisible by $8-3=5$, so $P(8)\equiv 7\equiv 2 \pmod 5$. Likewise $P(n)\equiv P(m)\pmod{n-m}$ for every pair of integers. This gives a quick way to rule out a candidate value without knowing the polynomial, and it also gives a way to construct polynomials: the linear polynomial $P(x)=7+m(x-3)$ with integer slope $m$ gives $P(8)=7+5m$, so every value in the right residue class mod $5$ is achieved by some integer-coefficient polynomial.

The same idea gives a divisibility test for roots. If $P(a)=0$ and $P(b)=0$ then $a-b\mid 0$ is automatic, so the principle is only useful when at least one value is nonzero. Its most common use is a contradiction: if $P(a)$ and $P(b)$ are given and $a-b$ does not divide their difference, then no integer-coefficient polynomial can take those values. Problems of this type ask whether a set of values is possible, and the answer is usually a single divisibility check.

For a polynomial with integer coefficients, also remember the parity trick: $P(x)\equiv P(x+2)\pmod 2$ and more generally $P(a)\equiv P(a\bmod m)\pmod m$ for any modulus $m$. When $x$ ranges over residues, $P(x)$ takes only the values allowed by its residue mod $m$. Combining the difference rule and the residue rule usually settles the question, so check both before doing heavy algebra.

• For integer-coefficient $P$ and integers $a,b$: $(a-b)\mid P(a)-P(b)$
• Equivalently $P(a)\equiv P(b)\pmod{a-b}$, so the residue of $P(a)$ mod $a-b$ is fixed
• For integer $P$ and modulus $m$, $P(a)\bmod m$ depends only on $a\bmod m$

• Claiming $P(a)-P(b)$ is divisible by $a+b$ or by $a^2-b^2$ instead of $a-b$.
• Using the principle with rational coefficients, where $(a-b)\mid P(a)-P(b)$ can fail.

6.Finite differences and the degree of a polynomial

Given values of a polynomial at equally spaced points $x=0,1,2,\dots$, the finite differences organise the information. Let $\Delta P(x)=P(x+1)-P(x)$, and write $\Delta^k$ for $k$ iterations. The key fact is that if $P$ has degree $d$, then $\Delta P$ has degree $d-1$, because the leading term $cx^d$ contributes $c\,d\,x^{d-1}$ to the difference and lower terms contribute less. Iterating, $\Delta^d P$ is a constant, and $\Delta^{d+1}P=0$. So a polynomial of degree $d$ is detected by the first difference column that becomes constant.

The leading coefficient is also read off. Since $\Delta^d(x^d)=d!$, a polynomial of degree $d$ with leading coefficient $c$ has constant $d$-th difference $c\cdot d!$. The lower difference columns give the polynomial itself, through Newton's forward-difference formula $P(x)=\sum_{k=0}^{d}\binom{x}{k}\Delta^kP(0)$. This formula is the discrete analogue of a Taylor expansion, and it lets you extrapolate: to find $P(n)$ you simply compute the next entries of each difference column.

Here is a practical method for a contest. Write the values in a row, subtract adjacent entries to get the first difference row, and repeat. When a row becomes constant, you know the degree and the constant. To extend the sequence, add the constant back up through the rows. This works without ever writing down the polynomial, which is valuable when the coefficients are messy but the values are clean.

A caution: finite differences characterise polynomials only on a lattice of equally spaced integers. If the points are not equally spaced, use Lagrange interpolation instead. Also, a sequence whose $d$-th difference is constant is a polynomial of degree $d$ in $n$ on the integers, but its values may not be polynomial at non-integers, so keep track of the domain in the problem.

• $\Delta^k$ of a degree-$d$ polynomial is a polynomial of degree $d-k$, so $\Delta^dP$ is constant
• $\Delta^d(x^d)=d!$, so a leading coefficient $c$ gives constant $d$-th difference $c\cdot d!$
• Newton's forward formula: $P(x)=\sum_{k=0}^{d}\binom{x}{k}\Delta^kP(0)$

• Stopping at a nonconstant difference row, or concluding the degree from one row only.
• Forgetting the factor $d!$ when reading the leading coefficient from the constant $d$-th difference.

7.Lagrange interpolation and uniqueness

Suppose you know the values of a polynomial at $n+1$ distinct points $x_0,\dots,x_n$. Then there is exactly one polynomial of degree at most $n$ taking those values. Existence follows from Lagrange's formula, and uniqueness follows from the factor theorem: if two such polynomials agreed at $n+1$ points, their difference would have degree at most $n$ and at least $n+1$ roots, so it would be zero. This is the sharpest use of the root-counting idea from the first concept.

The Lagrange basis polynomials make the existence explicit. Define $\ell_i(x)=\prod_{j\ne i}\frac{x-x_j}{x_i-x_j}$. Each $\ell_i$ has degree $n$, equals $1$ at $x_i$, and equals $0$ at every other node, because the product contains the factor $x-x_j$ for each $j\ne i$. The interpolating polynomial is then $P(x)=\sum_i y_i\,\ell_i(x)$, where $y_i=P(x_i)$. Evaluating at a node picks out exactly one term, which is why the formula reproduces the given data.

For contest work you rarely need the full expansion. Usually you need one value, $P(c)$, at a new point. Plugging $c$ into the formula gives $P(c)=\sum_i y_i\prod_{j\ne i}\frac{c-x_j}{x_i-x_j}$, a finite sum of fractions. Choose nodes that make most basis polynomials vanish, for example $x=0,1,2$ or a set containing your target point's neighbours, to keep the arithmetic small.

A powerful method uses uniqueness directly. If $P$ has degree at most $n$ and you know $n+1$ values, compare $P$ with a simpler polynomial $R$ that matches the same values, for instance a polynomial of the form $x^2$ or a product of linear factors. Their difference vanishes at all nodes, so it is a multiple of $\prod(x-x_j)$. Once you see that structure, the unknown part of $P$ has a single free multiplier, which a leading-coefficient condition pins down. The example below uses exactly this, and it avoids expanding any Lagrange sum.

Lagrange interpolation also explains why values determine polynomials so tightly. A polynomial of degree at most $5$ is fixed by six values, so a problem that supplies six values and asks for a seventh is really asking for the next value of a unique object, which the formula computes directly.

• A polynomial of degree at most $n$ is determined by its values at $n+1$ distinct points
• Lagrange basis: $\ell_i(x)=\prod_{j\ne i}\frac{x-x_j}{x_i-x_j}$, with $\ell_i(x_k)=\delta_{ik}$
• If $P$ and $R$ agree at $n+1$ points and both have degree $\le n$, then $P=R$

• Forgetting that the interpolant must have degree at most $n$, so the extra freedom is only the multiple $\prod(x-x_j)$.
• Mixing up node values and coefficients: the formula uses the values $y_i=P(x_i)$, not the coefficients of $P$.

8.Transforming roots by substitution

Often a problem gives a polynomial whose roots are $r_1,\dots,r_n$ and asks about a transformed set, such as $r_i^2$, $1/r_i$, or $r_i+c$. You can answer the question with Vieta alone, but it is cleaner to build a new polynomial whose roots are the transformed values, then read off the answer. The general principle is this: if $P(r)=0$ and $y=f(r)$, then the polynomial $Q(y)$ with roots $f(r_i)$ is the one obtained by eliminating $r$ from the system $y=f(r)$ and $P(r)=0$.

Three transformations are routine. For a shift $y=r+c$, substitute $r=y-c$ into $P$, so $Q(y)=P(y-c)$. For reciprocals $y=1/r$, the polynomial with roots $1/r_i$ is the reversal of the coefficients, $y^nP(1/y)$, up to scaling by the constant term. For squares, use the identity $P(x)P(-x)$, which is an even polynomial in $x$. Its roots are $\pm r_i$, so writing it as $Q(x^2)$ gives a polynomial whose roots are $r_i^2$. The sign is fixed by matching the leading coefficient.

The squares method is the one students find surprising, so here is the reason. The roots of $P(x)P(-x)$ are the $r_i$ together with the $-r_i$. Every such root has $x^2=r_i^2$, so $P(x)P(-x)=\pm Q(x^2)$, where $Q$ has roots $r_i^2$. The even polynomial $P(x)P(-x)$ is the device that removes the sign ambiguity: its odd part cancels.

Once the new polynomial $Q$ is known, its coefficients give symmetric sums of the new roots through Vieta. For example, if $P(x)=x^3-2x+5$, then $P(x)P(-x)=25-(x^3-2x)^2$, which equals $-(x^6-4x^4+4x^2-25)$. So the squared roots satisfy $y^3-4y^2+4y-25=0$, and their sum is $4$. This is a faster route than computing $\sum r_i^2=e_1^2-2e_2$ by hand, although both must agree, and checking them against each other is a useful habit.

• Shift: if $y=r+c$, the polynomial with roots $y_i$ is $P(y-c)$
• Reciprocal: the polynomial with roots $1/r_i$ is $y^nP(1/y)$ up to a constant multiple
• Squares: if $P(x)P(-x)=\pm Q(x^2)$, then $Q$ has roots $r_i^2$

• Taking the reciprocal polynomial without reversing the coefficients, which gives the wrong roots.
• Dropping the sign when squaring roots, so the constant term of $Q$ is wrong.

9.Complex and conjugate roots

A polynomial with real coefficients may have complex roots, but they arrive in pairs. If $P$ has real coefficients and $P(a+bi)=0$, then taking complex conjugates gives $P(a-bi)=0$, because conjugation commutes with addition and multiplication and fixes real numbers. So non-real roots come in conjugate pairs $a\pm bi$. Each such pair contributes a real quadratic factor $(x-a)^2+b^2=x^2-2ax+(a^2+b^2)$ with real coefficients.

The consequence for counting is decisive. A real polynomial of odd degree has at least one real root, because non-real roots pair up and cannot account for an odd number of roots. A real cubic therefore has either three real roots or one real root together with a conjugate pair. A real quartic can have $0$, $2$ or $4$ real roots, but never exactly one or three. These parity facts often rule out a configuration before any computation.

For a quadratic with roots $a\pm bi$, the sum and product are real: the sum is $2a$ and the product is $a^2+b^2$. This is why Vieta can be applied to a real polynomial whose roots include a complex pair without any imaginary arithmetic. If you know one complex root $z$ of a real polynomial, then $\bar z$ is also a root, and the quadratic $(x-z)(x-\bar z)$ is a factor with real coefficients, which you can divide out.

The example below uses this structure in a cubic. Given one complex root, the conjugate root and the real root follow, and Vieta then determines the coefficients. Keep track of signs and remember that the product of the three roots of a monic cubic equals minus the constant term.

• For real $P$: $P(z)=0$ implies $P(\bar z)=0$, so non-real roots come in pairs $a\pm bi$
• $(x-(a+bi))(x-(a-bi))=x^2-2ax+(a^2+b^2)$ is a real quadratic factor
• A real polynomial of odd degree has at least one real root

• Assuming a real cubic with one complex root has its other roots arbitrary; the conjugate is forced.
• Taking the conjugate of a root but not the sign of the imaginary part in the quadratic factor.

10.Roots of unity (introductory)

A root of unity is a complex number $\omega$ with $\omega^n=1$. There are exactly $n$ of them, namely $\omega_k=e^{2\pi ik/n}$ for $k=0,1,\dots,n-1$, and they are the vertices of a regular $n$-gon inscribed in the unit circle. Because $x^n-1$ has degree $n$ and these $n$ roots are distinct, the factor theorem gives $x^n-1=\prod_{k=0}^{n-1}(x-\omega_k)$. This factorisation is the basic identity behind the whole subject.

From the factorisation you get several facts at once. Setting $x=0$ gives $\prod(-\omega_k)=-1$ for odd $n$, and comparing coefficients gives Vieta for the roots of unity. Taking the coefficient of $x^{n-1}$ shows that the sum of all $n$-th roots of unity is $0$ whenever $n\ge 2$. Also $1+\omega+\omega^2+\cdots+\omega^{n-1}=0$ for $\omega\ne 1$, which follows from the geometric series formula $\frac{\omega^n-1}{\omega-1}=0$.

A standard technique is to differentiate the factorisation. Taking the logarithmic derivative gives $\sum_k \frac{1}{x-\omega_k}=\frac{nx^{n-1}}{x^n-1}$. Evaluating at a convenient $x$ turns a sum over roots of unity into a closed form. This is the key idea for problems asking for $\sum 1/(2-\omega_k)$ or similar sums.

A second standard fact is that the powers of $\omega$ cycle. If $\omega^n=1$, then $\omega^k$ depends only on $k\bmod n$, so a sum $\sum_{k=0}^{N-1}\omega^{k}$ is a sum over complete cycles plus a partial cycle, and complete cycles contribute $0$. For a cube root $\omega=e^{2\pi i/3}$ the identity $1+\omega+\omega^2=0$ is the most common building block.

Roots of unity appear in problems about integer polynomials with symmetry, and about counting with filters. At this level the goal is to recognise the factorisation $x^n-1=\prod(x-\omega_k)$ and the vanishing of sums of roots of unity. Harder uses, such as cyclotomic polynomials, belong to a later stage.

• $x^n-1=\prod_{k=0}^{n-1}(x-\omega_k)$ with $\omega_k=e^{2\pi ik/n}$
• For $n\ge 2$: $\sum_{k=0}^{n-1}\omega_k=0$, and $1+\omega+\omega^2=0$ for $\omega=e^{2\pi i/3}$
• $\sum_{k=0}^{n-1}\frac{1}{x-\omega_k}=\frac{nx^{n-1}}{x^n-1}$

• Forgetting that the $n$ roots of unity are distinct, which breaks the factorisation $x^n-1=\prod(x-\omega_k)$ if you list them twice.
• Writing $\omega^k$ for $k\ge n$ without reducing modulo $n$.

Example 1: Squares of roots of a quadratic

The roots $r$ and $s$ of $x^2-7x+k=0$ satisfy $r^2+s^2=25$. What is $k$?

Example 2: Reciprocals of roots

The roots of $x^3-6x^2+11x-6=0$ are $r,s,t$. What is $\frac{1}{r}+\frac{1}{s}+\frac{1}{t}$?

Example 3: Remainder by a product of three linear factors

A polynomial $P(x)$ leaves remainder $2$ when divided by $x-1$, remainder $3$ when divided by $x-2$, and remainder $6$ when divided by $x-3$. What is the remainder when $P(x)$ is divided by $(x-1)(x-2)(x-3)$, evaluated at $x=4$?

Example 4: Shifted roots and a sum of reciprocals

Let $r,s,t$ be the roots of $x^3+2x-5=0$. What is $\frac{1}{r+1}+\frac{1}{s+1}+\frac{1}{t+1}$?

Example 5: A monic quartic from four values

$P(x)$ is a monic quartic polynomial with $P(1)=3$, $P(2)=5$, $P(3)=7$ and $P(4)=9$. What is $P(5)$?

Example 6: Sum over nontrivial seventh roots of unity

Let $\omega=e^{2\pi i/7}$. What is $\sum_{k=1}^{6}\frac{1}{1-\omega^k}$?