Algebra & Polynomials

Why it works. A monic polynomial with roots $r_1, \dots, r_n$ IS the product $(x - r_1)\cdots(x - r_n)$. Expanding, the $x^{n-1}$ term collects $-r_i$ once from each factor, the $x^{n-2}$ term collects every product of two roots, and the constant is the product of all the $-r_i$. Matching coefficients is the whole theorem, which is also how to rebuild the signs if you forget them: they alternate, starting with minus.

Why it works. Division gives $P(x) = (x - a)Q(x) + R$ with $R$ a constant, because the remainder has lower degree than $x - a$. Put $x = a$: the first term vanishes and $P(a) = R$. Dividing by a quadratic leaves a LINEAR remainder $cx + d$, and two evaluations pin down its two unknowns.

Spotting it. A product of the form $\prod(\text{root} + k)$ is the polynomial itself, evaluated at $-k$. No expansion into symmetric sums is needed.

Step 1. $P(x) = (x - a)(x - b)(x - c)$, so $P(-1) = (-1 - a)(-1 - b)(-1 - c) = -(a + 1)(b + 1)(c + 1)$.

Step 2. $P(-1) = -1 - 5 - 3 + 1 = -8$, so $(a + 1)(b + 1)(c + 1) = 8$.

Check by Vieta: the product expands to $abc + (ab + bc + ca) + (a + b + c) + 1 = -1 + 3 + 5 + 1 = 8$ ✓

Spotting it. The divisor is a quadratic, so the remainder is linear: $P(x) = (x - 1)(x - 2)Q(x) + cx + d$. The remainder theorem gives $P(1)$ and $P(2)$, which is exactly two equations.

Step 1. $x = 1$: $c + d = P(1) = 3$. $x = 2$: $2c + d = P(2) = 5$.

Step 2. Subtract: $c = 2$, then $d = 1$. The remainder is $2x + 1$.

Spotting it. Long division is out of the question. The divisor factors as $(x - 1)(x + 1)$, so evaluate at its two roots.

Step 1. Write $x^{51} + 2 = (x^2 - 1)Q(x) + cx + d$.

Step 2. $x = 1$: $c + d = 3$. $x = -1$: $-c + d = -1 + 2 = 1$.

Step 3. So $d = 2$, $c = 1$, and the remainder is $x + 2$.

Vieta answers questions about the roots without finding them, which is the point: for $x^2-7x+10$, $r_1^2+r_2^2 = (r_1+r_2)^2 - 2r_1r_2 = 49 - 20 = 29$ with no factoring at all. The identities worth knowing cold are $\sum r^2 = e_1^2 - 2e_2$ and $\sum \tfrac1r = \tfrac{e_2}{e_3}$ for a cubic.

It turns a two-variable equation into a divisor problem. $xy + 3x + 2y = 20$ becomes $(x+2)(y+3) = 26$, and the integer solutions come from the divisor pairs of $26$. The signal is a product term plus linear terms in the same equation — the missing constant is always the product of the two linear coefficients.

Step 1 — Vieta: $e_1 = 6$, $e_2 = 11$, $e_3 = 6$.

Step 2 — $\sum r^2 = e_1^2 - 2e_2 = 36 - 22 = 14$.

Step 3 — $\sum \tfrac1r = \tfrac{e_2}{e_3} = \tfrac{11}{6}$. Check: the roots are $1, 2, 3$, and $1 + 4 + 9 = 14$, $1 + \tfrac12 + \tfrac13 = \tfrac{11}{6}$ ✓

Spotting it. Clearing denominators gives a product term and two linear terms, $mn - 6m - 6n = 0$: the signal for Simon's trick.

Step 1. Add $36$ to both sides: $(m - 6)(n - 6) = 36$.

Step 2. Both factors must be positive. If both were negative, then $m - 6 \ge -5$ and $n - 6 \ge -5$ would make their product at most $25 < 36$.

Step 3. So $m - 6$ is any positive divisor of $36 = 2^2 3^2$, and $n - 6$ is the complementary divisor. There are $(2 + 1)(2 + 1) = 9$ divisors, hence 9 ordered pairs, from $(7, 42)$ to $(42, 7)$.

Spotting it. $r^3 + s^3$ is symmetric, so it is a polynomial in $r + s = 4$ and $rs = 1$.

Step 1. $(r + s)^3 = r^3 + s^3 + 3rs(r + s)$, so $r^3 + s^3 = (r + s)^3 - 3rs(r + s)$.

Step 2. $= 64 - 3 \cdot 1 \cdot 4 = 52$. The roots themselves are $2 \pm \sqrt3$, and nothing in the computation needed them.

Why it works. Every symmetric polynomial in $x$ and $y$ can be written in terms of $s$ and $p$ alone, so a symmetric system in $x, y$ becomes a system in $s, p$, usually of lower degree. Going back is Vieta: numbers with sum $s$ and product $p$ are the roots of $t^2 - st + p$. The discriminant $s^2 - 4p = (x - y)^2$ is a square of a real number, which is why it must be non-negative when $x$ and $y$ are real.

Spotting it. Swapping $x$ and $y$ changes nothing, so switch to $s$ and $p$; $s = 5$ is given.

Step 1. $x^3 + y^3 = s^3 - 3ps$ gives $35 = 125 - 15p$, so $p = 6$.

Step 2. $x$ and $y$ are the roots of $t^2 - 5t + 6 = (t - 2)(t - 3)$: the solutions are $(2, 3)$ and $(3, 2)$.

Check: $8 + 27 = 35$ ✓

Spotting it. Both equations are symmetric. In $s$ and $p$ they read $s + p = 11$ and $sp = 28$ (since $x^2y + xy^2 = xy(x + y)$): a sum and a product again, so $s$ and $p$ are themselves the roots of a quadratic.

Step 1. $t^2 - 11t + 28 = (t - 4)(t - 7)$, so $\{s, p\} = \{4, 7\}$.

Step 2. Case $s = 4$, $p = 7$: the reality test fails, since $s^2 = 16 < 28 = 4p$. No real $x$, $y$ come from this case.

Step 3. Case $s = 7$, $p = 4$: real, since $49 \ge 16$. Then $x^2 + y^2 = 49 - 8 = 41$, and this is the only possible value. (The numbers are $\tfrac{7 \pm \sqrt{33}}{2}$.)