$$(1+x)^5,\ (1-x^2)^6,\ (3x+x^3)^8$$

'Bi' means two terms in the bracket, one power outside. Reflex: Check the bracket has exactly two terms before applying any binomial tool.

$$(a+b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \cdots + \binom{n}{n-1}ab^{n-1} + \binom{n}{n}b^n$$

$\binom{n}{r}$ = "n choose r", the binomial coefficients in the expansion. Reflex: Never compute $\binom{n}{r}$ by hand — press ${}^nC_r$ on your calculator.

$$(a+b)^n = \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1}b^1 + \binom{n}{2}a^{n-2}b^2 + \cdots$$

Here $a$ is the first term in the bracket, $b$ the second, $n$ the power outside. Watch $a$'s power fall ($n, n{-}1, n{-}2, \ldots$) while $b$'s rises ($0, 1, 2, \ldots$) — every pair adds up to $n$, across $(n+1)$ terms.

Reflex: Track $a$'s power falling and $b$'s power rising — the two powers always sum to $n$.

$$(-2x)^3 = -8x^3$$

The power hits the coefficient AND the sign AND the $x$, not just the $x$.

Reflex: Write $(-2x)^r$ with brackets every time; odd $r$ keeps the minus, even $r$ kills it.

Want one coefficient? Truncate every factor at that power — no full expansion needed. Multiply out only the pairs whose powers add up to the target, then stop.

Reflex: Want $x^2$? Stop every factor at $x^2$, collect only powers adding to 2.

A constant term has no $x$ in it. When two brackets multiply, you get a constant wherever the $x$'s cancel out:

$$\begin{aligned}&\text{const} \times \text{const}\\&\tfrac{1}{x} \times x\\&\tfrac{1}{x^2} \times x^2\\&\ldots\end{aligned}$$

Multiply each such pair, then add them all up — that total is the constant term.

Reflex: Look for pairs whose $x$'s cancel out — those pairs make the constant term.

$$T_{r+1} = \binom{n}{r}a^{n-r}b^r$$

Identify $n,a,b$, combine powers of $x$, equate to the target power, solve for $r$.

Reflex: The counting starts at $r = 0$, so the term number is always $r+1$: $r = 3$ is the 4th term. A fractional $r$ means that term doesn't exist.

$$\begin{aligned}\binom{n}{1} &= n\\\binom{n}{2} &= \frac{n(n-1)}{1\times 2}\\\binom{n}{3} &= \frac{n(n-1)(n-2)}{1\times 2\times 3}\\\binom{n}{4} &= \frac{n(n-1)(n-2)(n-3)}{1\times 2\times 3\times 4}\end{aligned}$$

Calculator's useless when $n$ is unknown — turn coefficient conditions into a polynomial equation in $n$.

Reflex: Solve, then reject $n=0$ and fractions — $n$ must be a positive integer.

$$\left(-\frac{1}{3x^3}\right)^r = \left(-\frac{1}{3}\right)^r x^{-3r}, \quad x^{12-r} \cdot x^{-3r} = x^{12-4r}$$

Pull out numbers and sign first, leaving pure $x$-powers. Only after adding exponents can you equate the power.

Reflex: Add exponents when multiplying powers of $x$; let the sign ride along as $(-1)^r$ or $\left(-\frac{1}{3}\right)^r$.

Write coefficients with $\binom{n}{r}$ and constants (no $x$) → fraction → cancel repeated powers (indices laws) → cross multiply, swap $\binom{n}{r}$ for its formula in $n$ → solve, reject roots that aren't positive integers.

Reflex: Ratio → fraction → cancel powers → cross multiply → solve → reject non-integers.

Match the given form to the target value, solve for $x$, substitute into the truncated expansion, then round.

Tiny $x$ (like 0.02) kills later terms; don't assume this holds if $x$ is larger.

Reflex: Solve for x, substitute, and check the first omitted term is negligible before rounding.