$$ax^n + bx^{n-1} + cx^{n-2} + \dots + dx + e$$

Degree is the highest whole-number power of $x$; a constant alone has degree $0$. Reflex: Spot the fakes—$\frac{1}{x}$ or $\sqrt{x}$ means it's NOT a polynomial.

$$\begin{gathered}\text{Dividend} = \text{Divisor} \times \text{Quotient}\\[2pt] {} + \text{Remainder}\end{gathered}$$

In symbols: $f(x) = \text{divisor} \times Q(x) + \text{remainder}$. The remainder is one degree below the divisor: linear divisor → constant remainder; quadratic divisor → $ax+b$. You never need to find $Q(x)$ — sub a root of the divisor and the whole $\text{divisor} \times Q(x)$ term becomes $0$, leaving only the remainder. If the divisor has no real roots, compare coefficients instead. Reflex: Write $f(x) = \text{divisor} \times Q(x) + \text{remainder}$, then sub roots of the divisor — the $Q(x)$ term becomes $0$.

$$\begin{gathered}f(x) \div (x-a):\\[1pt] \text{Remainder} = f(a)\\[6pt] f(x) \div (ax-b):\\[1pt] \text{Remainder} = f\!\left(\tfrac{b}{a}\right)\end{gathered}$$

Subbing kills the $Q(x)$ term, leaving just the remainder — not the quotient.

Reflex: Set divisor $=0$ first to find what to sub in.

$$\begin{gathered}(x-a) \text{ is a factor of } f(x)\\[2pt] \iff f(a) = 0\end{gathered}$$

Check a factor: compute $f(a)$, see if it's $0$. Find unknowns: set $f(a)=0$ and solve. For $(ax-b)$, test $f(b/a)$. Reflex: Spot 'exactly divisible by $x-2$' and instantly substitute $f(2)=0$.

$$f(x)=(x-a)(bx^2+cx+d)$$

Trial small values till $f(a)=0$, divide out $(x-a)$, then factorize the quadratic as usual. Match $b$ to $x^3$, $d$ to the constant, find $c$ from a middle term. Reflex: Guess one root to open the cubic into a familiar quadratic.

$$\begin{gathered} a^3+b^3 = (a+b)(a^2-ab+b^2) \\ a^3-b^3 = (a-b)(a^2+ab+b^2) \end{gathered}$$

Middle sign is opposite the first bracket's; coefficient is always $1$, never $2$ — second bracket won't factor further.

Reflex: Same sign first, opposite in the middle, plus $b^2$ — not in the formula sheet.

Identity ($\equiv$) holds for all $x$: sub values that zero out brackets to kill unknowns, then match coefficients of each power left over.

Reflex: Sub the roots first, then compare the highest power for what's left.

$$f(x) = \text{divisor} \times Q(x) + (ax+b)$$

Remainder is $ax+b$ (degree = divisor's − 1). Factors? Sub roots to zero out $Q(x)$. No real roots? Compare coefficients. Reflex: Pick roots of the divisor so $Q(x)$ vanishes — never solve for $Q(x)$.

$$f(x) = -2(x+1)(x-2)(x+k)$$

Each root $r$ gives factor $(x-r)$; short on roots for the degree? Top up with $(ax^2+bx+c)$. Constant term = product of brackets' constants. Reflex: Wrap the coefficient around everything — never tack it on the end.