Polynômes

$P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$, $a_n\neq 0$

Termone $a_kx^k$·Coefficientthe number $a_k$·Constant$a_0=P(0)$

Leading$a_n$, the coefficient of the highest power·Monic$a_n=1$

Degree$\deg P=n$, the highest exponent·several variables: add the exponents, $\deg(x^3y^2)=5$

Standard formterms written once each, powers descending, like terms already merged

Like termssame variables to the same powers: $5x^2y$ and $-2x^2y$·only these may be added

Zero poly.$P\equiv 0$, degree left undefined·Completeno power between $n$ and $0$ is missing

$\deg(PQ)=\deg P+\deg Q$·$\deg(P+Q)\leqslant\max(\deg P,\deg Q)$, it can drop

By number of terms

Monôme one term: $7x^3$, $-\tfrac12 ab^2$·a product of a number and powers, nothing added

Binôme two terms: $x^2-9$, $3x+1$·Trinôme three: $x^2+5x+6$

Polynôme any finite sum of monomials·four or more terms get no special name

By degree$0$ constant $c$·$1$ linear $ax+b$·$2$ quadratic·$3$ cubic

$4$ quartic·$5$ quintic·beyond that, degree $n$

Alsohomogeneous: every term the same degree, $x^3+x^2y+y^3$

symmetric: unchanged when variables swap·palindromic: coefficients read the same backwards

Ruleadd coefficients of like terms, keep the variable part untouched

$a_kx^k+b_kx^k=(a_k+b_k)x^k$·$3x^2+5x^2=8x^2$, and $3x^2+5x^3$ stays as it is

Subtractthe minus reaches every term inside the bracket

$(4x^2-3x+1)-(x^2+2x-5)=3x^2-5x+6$

Methoddrop brackets with the right signs, group like powers, add, write in descending order

Degreessame degree with opposite leading coefficients: the top term cancels and the degree drops

$(x^3+2x)-(x^3-x)=3x$·degree fell from $3$ to $1$

Structure$P+Q=Q+P$·$P-Q=P+(-Q)$·the zero polynomial is neutral

Monomials$(ax^m)(bx^k)=ab\,x^{m+k}$·multiply numbers, add exponents

By a monomial$a(b+c+d)=ab+ac+ad$, every term gets hit

Two binomials$(a+b)(c+d)=ac+ad+bc+bd$ (first, outer, inner, last)

General$\left(\displaystyle\sum_i a_ix^i\right)\!\left(\sum_j b_jx^j\right)=\sum_k\left(\sum_{i+j=k}a_ib_j\right)x^k$

each coefficient of the product is one diagonal of the multiplication table

Powers$P^n$ by the binomial theorem when $P$ is a binomial, otherwise repeated multiplication

Check$\deg$ adds·leading coefficients multiply·constant terms multiply

$P(1)$ is the sum of all coefficients, a fast sanity test after expanding

By a monomial$\dfrac{a+b}{m}=\dfrac{a}{m}+\dfrac{b}{m}$·divide every term, subtract exponents

Euclideanfor $D\neq0$ there are unique $Q,R$ with $P=Q\cdot D+R$, $\deg R<\deg D$

Long divisiondivide leading by leading, multiply back, subtract, bring down, repeat

Hornerdividing by $(x-c)$: bring down $a_n$, then each next entry is $a_k+c\cdot(\text{previous})$

the last entry is $P(c)$, the ones before it are the coefficients of $Q$

Remainder$P(x)=(x-c)Q(x)+P(c)$·so $R=P(c)$ for a linear divisor

Exact$D\mid P\iff R=0$·then $\deg Q=\deg P-\deg D$

Improper$\deg P\geqslant\deg D$: divide first, then split off $\dfrac{R}{D}$

$(a\pm b)^2=a^2\pm 2ab+b^2$·$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

$(a\pm b)^3=a^3\pm 3a^2b+3ab^2\pm b^3$·$(a+b)(a-b)=a^2-b^2$

$a^3+b^3=(a+b)(a^2-ab+b^2)$·$a^3-b^3=(a-b)(a^2+ab+b^2)$

$a^n-b^n=(a-b)\!\left(a^{n-1}+a^{n-2}b+\cdots+b^{n-1}\right)$ for every $n$

$a^n+b^n=(a+b)\!\left(a^{n-1}-a^{n-2}b+\cdots+b^{n-1}\right)$ for odd $n$ only

$x^2+(p+q)x+pq=(x+p)(x+q)$·$a^2+2ab+b^2$ and $a^2-2ab+b^2$ are the perfect squares

Binomial$(a+b)^n=\displaystyle\sum_{k=0}^{n}\binom{n}{k}a^{\,n-k}b^{\,k}$·coefficients from Pascal: 1 4 6 4 1

Careful$a^2+b^2$ has no factorisation over $\mathbb{R}$; over $\mathbb{C}$ it is $(a+bi)(a-bi)$

Step 0always pull the greatest common factor out first: $6x^3-9x^2=3x^2(2x-3)$

2 termstry $a^2-b^2$, then $a^3\pm b^3$, then $a^n\pm b^n$

3 termsperfect square?·$x^2+bx+c$: find $p+q=b$ and $pq=c$, then $(x+p)(x+q)$

$ax^2+bx+c$: find $p\cdot q=ac$ with $p+q=b$, split the middle term, group

4 termsgroup in pairs: $ax+ay+bx+by=a(x+y)+b(x+y)=(a+b)(x+y)$

Substitution$x^4+5x^2+4$ with $t=x^2$·$(x^2+x)^2-2(x^2+x)-8$ with $t=x^2+x$

Add and subtract$x^4+4=x^4+4x^2+4-4x^2=(x^2+2)^2-(2x)^2$

Stop whenevery factor is irreducible over the set you are working in

Factor thm.$P(c)=0\iff (x-c)\mid P$·one root gives one linear factor

Rational rootsfor integer coefficients, any rational root $\dfrac{p}{q}$ has $p\mid a_0$, $q\mid a_n$

so test the finite list $\pm\dfrac{\text{divisors of }a_0}{\text{divisors of }a_n}$, nothing else can work

Peelfound $c$: divide by $(x-c)$ with Horner, then factor the smaller quotient the same way

Multiplicity$c$ has multiplicity $m$ when $(x-c)^m\mid P$ but $(x-c)^{m+1}\nmid P$

Quadratic left$D=b^2-4ac$·$D\geqslant 0$: split further·$D<0$: irreducible over $\mathbb{R}$

Countthe multiplicities of all roots add up to $\deg P$ once you are over $\mathbb{C}$

Fast checks$P(1)=0$ iff coefficients sum to $0$·$P(-1)=0$ iff the alternating sum is $0$

Form$\dfrac{P(x)}{Q(x)}$ with $Q\neq0$·state the excluded values before touching anything

Simplifyfactor top and bottom, then cancel common factors, never single terms

$\dfrac{x^2-9}{x^2+3x}=\dfrac{(x-3)(x+3)}{x(x+3)}=\dfrac{x-3}{x}$, valid for $x\neq0,-3$

Multiply$\frac{A}{B}\cdot\frac{C}{D}=\frac{AC}{BD}$·Divideflip the second one·Stacked$\frac{A/B}{C/D}=\frac{AD}{BC}$

Addfactor every denominator, take the LCD, raise each fraction, then add numerators

Improper$\deg P\geqslant\deg Q$: divide first, $\frac{P}{Q}=S+\frac{R}{Q}$

Partial fractions$\dfrac{R}{(x-a)^m(x^2+px+q)^k}=\displaystyle\sum\frac{A_i}{(x-a)^i}+\sum\frac{B_jx+C_j}{(x^2+px+q)^j}$

Same thing$c$ is a root $\iff$ $(x-c)$ is a factor $\iff$ the graph meets the axis at $c$

Countat most $n$ real roots·at most $n-1$ turning points·exactly $n$ roots over $\mathbb{C}$

Multiplicityodd: the graph crosses·even: it touches and turns back

higher multiplicity flattens the curve near the root

Ends$n$ even, $a_n>0$: both arms up·$n$ even, $a_n<0$: both down

$n$ odd, $a_n>0$: down then up·$n$ odd, $a_n<0$: up then down

Sign change$P(a)P(b)<0$ guarantees a root between $a$ and $b$

Descartessign changes in $P(x)$ bound the positive roots, in $P(-x)$ the negative ones, each dropping by $2$

Quadratic$x_1+x_2=-\dfrac{b}{a}$·$x_1x_2=\dfrac{c}{a}$

Cubic$\sum x_i=-\dfrac{b}{a}$·$\sum_{i<j} x_ix_j=\dfrac{c}{a}$·$x_1x_2x_3=-\dfrac{d}{a}$

General$e_k=(-1)^k\dfrac{a_{n-k}}{a_n}$, where $e_k$ is the sum of all products of $k$ distinct roots

Rebuildknowing $s=x_1+x_2$ and $p=x_1x_2$ gives $x^2-sx+p=0$

Power sums$p_1=e_1$·$p_2=e_1^2-2e_2$·$p_3=e_1^3-3e_1e_2+3e_3$ (Newton)

Symmetricany symmetric expression in the roots is a polynomial in $e_1,\ldots,e_n$

so $x_1^2+x_2^2$, $\dfrac{1}{x_1}+\dfrac{1}{x_2}$, $(x_1-x_2)^2$ all follow from the coefficients alone

$(x_1-x_2)^2=e_1^2-4e_2=\dfrac{D}{a^2}$, the discriminant in disguise

Euclid$\gcd(P,Q)=\gcd(Q,P\bmod Q)$, stop when the remainder is $0$

Bézout$UP+VQ=\gcd(P,Q)$ for some polynomials $U,V$

Coprime$\gcd=1$·then $P\mid QR\Rightarrow P\mid R$·$\gcd\cdot\operatorname{lcm}=PQ$ up to a constant

Derivative$\left(x^k\right)'=kx^{k-1}$·$(PQ)'=P'Q+PQ'$

Repeated roots$c$ repeats $\iff P(c)=P'(c)=0$·$\dfrac{P}{\gcd(P,P')}$ has the same roots, all simple

Irreducibleno factorisation into two factors of lower degree, over the given set

degree $2$ or $3$: irreducible over $K$ $\iff$ it has no root in $K$·degree $4$ can factor with no roots

Testsrational root test, Eisenstein, reduction mod $p$

Biquadratic$ax^4+bx^2+c$: set $t=x^2$, solve, then take both square roots back

Repeat a block$(x^2+3x)^2-8(x^2+3x)+12$: set $t=x^2+3x$

Palindromicdivide by $x^{n/2}$, set $t=x+1/x$, then $x^2+1/x^2=t^2-2$

Shift$x=y-\frac{a_{n-1}}{na_n}$ kills the $x^{n-1}$ term; a cubic becomes $y^3+py+q$

Square$ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^{\!2}+c-\frac{b^2}{4a}$

Worth knowing$x^n-1=(x-1)(x^{n-1}+\cdots+x+1)$

$x^4+x^2+1=(x^2+x+1)(x^2-x+1)$·homogeneous in $x,y$: set $t=x/y$

Unknown coefficientsassume $(x^2+ax+b)(x^2+cx+d)$, expand, match term by term

Cancellingonly whole factors go: $\dfrac{x^2+x}{x}=x+1$ after factoring, and $\dfrac{x+2}{2}\neq x$

Domaincancelling $(x+3)$ removes the symbol, not the restriction $x\neq-3$

Minus sign$-(x^2-3x+4)=-x^2+3x-4$, every term flips

Squares$(a+b)^2\neq a^2+b^2$·$\sqrt{a^2+b^2}\neq a+b$

GCFskipping it leaves an answer that is factored, but not factored completely

Over what$x^2-2$ is irreducible over $\mathbb{Q}$ and reducible over $\mathbb{R}$; say which set you mean

Dividingdividing an equation by $x$ throws away the root $x=0$; factor instead

Degrees$\deg(P+Q)$ can be smaller than both·and $P(1)$ is a free check on any expansion