Polynomials are one of the most frequent topics in the YÖS MAT-2 section. Typical questions ask you to: determine when an expression is a polynomial (conditions on exponents), recover $P(x)$ from a shifted argument such as $P(x-3)$ or $P(2-x)$, compute special values like $P(1)$ (sum of coefficients) and $P(0)$ (constant term), find remainders with Bezout's theorem, divide by a quadratic like $x^2-1$, use equality of polynomials to compare coefficients, factor cubics to find integer roots, and apply degree rules such as $der[P\cdot Q]=der\,P+der\,Q$. Almost every exam problem reduces to a small set of rules covered below — learn them and practice substitutions.
A polynomial in one variable is an expression $$P(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0,\qquad a_n\neq 0,$$ where the coefficients $a_i$ are real numbers and every exponent of $x$ is a natural number (a non-negative integer: $0,1,2,\dots$). The number $n$ is the degree of the polynomial, written $\deg P$ (in Turkish sources $der[P(x)]$); $a_n$ is the leading coefficient and $a_0$ is the constant term. Key condition: an expression is a polynomial only if it contains no negative exponents (like $x^{-2}$), no fractional exponents (like $x^{1/2}=\sqrt{x}$), and no variable in a denominator. A nonzero constant is a polynomial of degree $0$.
Problem. $P(x)=x^{\frac{12}{m}}+x^{m-4}-3x+1$ is a polynomial. Find the sum of all possible values of $m$.
Solution. Every exponent must be a natural number.
1) $\frac{12}{m}\in\mathbb{N}$, so $m$ must be a positive divisor of $12$: $m\in\{1,2,3,4,6,12\}$.2) $m-4\in\mathbb{N}$, so $m-4\ge 0$, that is $m\ge 4$.
Both conditions together give $m\in\{4,6,12\}$. Check: $\frac{12}{4}=3$, $\frac{12}{6}=2$, $\frac{12}{12}=1$ and $m-4=0,\,2,\,8$ — all natural numbers.
Answer: $4+6+12=22$.
For $P(x)=a_nx^n+\dots+a_1x+a_0$:$$P(1)=a_n+a_{n-1}+\dots+a_1+a_0 \quad\text{(sum of all coefficients)}$$$$P(0)=a_0 \quad\text{(constant term)}$$$$\frac{P(1)+P(-1)}{2}=\text{sum of coefficients of even powers},\qquad \frac{P(1)-P(-1)}{2}=\text{sum of coefficients of odd powers}.$$If the polynomial is given with a shifted argument, make the argument equal to the needed value: for $P(x-3)$ the sum of coefficients of $P$ is obtained at $x=4$ (then $x-3=1$), and the constant term at $x=3$ (then $x-3=0$).
Example: $P(x)=3x^{11}-4x^7+kx$ and $P(1)=0$. Then $3-4+k=0$, so $k=1$.
To recover $P(x)$ when $P$ is given at a shifted argument, introduce a new variable equal to that argument.
Example: $P(x-3)=x^2-4x+3$. Let $t=x-3$, so $x=t+3$. Substitute:$$P(t)=(t+3)^2-4(t+3)+3=t^2+6t+9-4t-12+3=t^2+2t.$$Hence $P(x)=x^2+2x$.
The same works for decreasing arguments: for $P(2-x)$ let $t=2-x$, so $x=2-t$. Alternative shortcut: if only one value like $P(5)$ is needed, do not find the whole polynomial — just choose $x$ so that the argument equals $5$ (for $P(x-3)$ take $x=8$).
For nonzero polynomials $P$ and $Q$:$$der[P(x)\cdot Q(x)]=der[P(x)]+der[Q(x)]$$$$der\big[P\big(Q(x)\big)\big]=der[P(x)]\cdot der[Q(x)]$$$$der\big[P(x)^k\big]=k\cdot der[P(x)],\qquad der[P(x)\pm Q(x)]\le\max\big(der[P(x)],\,der[Q(x)]\big).$$Note that $der[P(x-3)]=der[P(x)]$: shifting the argument does not change the degree.
Parametric example: if $der[P(x)\cdot Q(x)]=7n+3$ and $der[P(x)]=3n+1$, then $$der[Q(x)]=(7n+3)-(3n+1)=4n+2.$$
Dividing $P(x)$ by a nonzero polynomial $B(x)$ gives a unique quotient $Q(x)$ and remainder $R(x)$: $$P(x)=B(x)\cdot Q(x)+R(x),\qquad der[R(x)]<der[B(x)].$$Bezout's theorem (remainder theorem): the remainder of $P(x)$ divided by $x-a$ equals $P(a)$. In particular, the remainder modulo $x+3$ is $P(-3)$, because $x+3=x-(-3)$.
Divisibility criterion: $P(x)$ is divisible by $x-a$ if and only if $P(a)=0$, i.e. $a$ is a root of $P$. So «$P(x)$ is divisible by $x-2$» means exactly $P(2)=0$.
If the polynomial is given at shifted arguments (e.g. $P(x-2)$), first choose $x$ so that the argument becomes the number you need: to get $P(3)$ from $P(x-2)$, take $x=5$.
When dividing by a polynomial of degree $2$, the remainder has degree at most $1$, so it can be written as $R(x)=ax+b$:$$P(x)=(x^2-1)\cdot Q(x)+ax+b.$$Since $x^2-1=(x-1)(x+1)$, substitute the roots $x=1$ and $x=-1$ (the term with $Q$ vanishes):$$P(1)=a+b,\qquad P(-1)=-a+b.$$Solve this system for $a$ and $b$.
Mini-example: the remainder of $P(x)$ modulo $x-1$ is $5$ and modulo $x+1$ is $1$. Then $a+b=5$ and $-a+b=1$, so $b=3$, $a=2$, and the remainder modulo $x^2-1$ is $2x+3$. The same method works for $x^2-4=(x-2)(x+2)$, $x^2-5x+6=(x-2)(x-3)$, etc.
Two polynomials are equal (as an identity, for every $x$) if and only if the coefficients of like powers of $x$ coincide. If $P(x)=Q(x)$ for all $x$, compare the coefficients of $x^n$, $x^{n-1}$, ..., and the constant terms one by one. You may also substitute convenient values of $x$ to produce equations.
Example: $(x^3+6)P(x)+6x=ax^3+2ax+b+3$ holds for all $x$. The right side has degree $3$ and the left side contains $x^3\cdot P(x)$, so $P(x)=c$ must be a constant. The left side becomes $cx^3+6x+6c$. Matching coefficients: $x^3$ gives $c=a$; $x^1$ gives $6=2a$, so $a=3$ and $c=3$; the constant term gives $6c=b+3$, so $b=15$. Hence $a+b=18$.
Integer root test: if a polynomial with integer coefficients has an integer root, that root divides the constant term. So to find integer roots of $x^3+3x^2-4x-12=0$, test the divisors of $12$: $\pm1,\pm2,\pm3,\pm4,\pm6,\pm12$.
Grouping is often faster:$$x^3+3x^2-4x-12=x^2(x+3)-4(x+3)=(x+3)(x^2-4)=(x+3)(x-2)(x+2).$$The roots are $x=-3$, $x=2$, $x=-2$; the product of the integer roots is $(-3)\cdot 2\cdot(-2)=12$.
General scheme: find one root $a$ (by testing divisors of the constant term), then divide by $x-a$ (long division or Horner's scheme) and factor the remaining quadratic.
$$(a\pm b)^2=a^2\pm 2ab+b^2,\qquad a^2-b^2=(a-b)(a+b)$$$$(a+b)^3=a^3+3a^2b+3ab^2+b^3,\qquad (a-b)^3=a^3-3a^2b+3ab^2-b^3$$$$a^3+b^3=(a+b)(a^2-ab+b^2),\qquad a^3-b^3=(a-b)(a^2+ab+b^2)$$Key pattern for polynomial problems: recognize a perfect cube inside a polynomial,$$x^3+3x^2+3x+1=(x+1)^3,\qquad x^3-3x^2+3x-1=(x-1)^3,$$ and a perfect square $x^2-6x+9=(x-3)^2$. This turns evaluating $P$ at irrational points like $\sqrt[3]{10}-1$ into a one-line computation.
Problem. $P(2-x)=x^2+2x+1$. Find $P(x)$ and compute $P(5)$.
Solution. Let $t=2-x$, so $x=2-t$. Substitute:$$P(t)=(2-t)^2+2(2-t)+1=4-4t+t^2+4-2t+1=t^2-6t+9=(t-3)^2.$$Hence $P(x)=(x-3)^2$ and $P(5)=(5-3)^2=4$.
Quick check without finding $P$: to get $P(5)$ directly, choose $x$ with $2-x=5$, i.e. $x=-3$. Then $P(5)=(-3)^2+2\cdot(-3)+1=9-6+1=4$. The answers agree.
Answer: $P(x)=(x-3)^2$, $P(5)=4$.
Problem. For every $x$, $P(x-2)+P(x+2)=4x+2$. Find the remainder when $P(x)$ is divided by $x-3$.
Solution. The right side has degree $1$, and shifting an argument does not change the degree, so $P$ must be linear: $P(x)=mx+c$. Then$$P(x-2)+P(x+2)=m(x-2)+c+m(x+2)+c=2mx+2c.$$Comparing with $4x+2$: $2m=4$ and $2c=2$, so $m=2$, $c=1$ and $P(x)=2x+1$.
By Bezout's theorem the remainder modulo $x-3$ equals $P(3)=2\cdot 3+1=7$.
Useful shortcut for such identities: substituting a well-chosen $x$ gives values instantly. For example, $x=2$ yields $P(0)+P(4)=10$ without finding $P$ at all.
Answer: $7$.
Problem. $P(x)=x^3+3x^2+3x+1$. Compute $P\big(\sqrt[3]{10}-1\big)$.
Solution. Recognize the short multiplication formula $(a+b)^3=a^3+3a^2b+3ab^2+b^3$ with $a=x$, $b=1$:$$P(x)=x^3+3x^2+3x+1=(x+1)^3.$$Now substitute $x=\sqrt[3]{10}-1$:$$P\big(\sqrt[3]{10}-1\big)=\big(\sqrt[3]{10}-1+1\big)^3=\big(\sqrt[3]{10}\big)^3=10.$$Moral: before substituting an ugly number, always try to collapse the polynomial into a power of a binomial — YÖS problems are built so that the irrational parts cancel.
Answer: $10$.
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