Unit B2: Polynomial
Functions单元 B2：多项式函数
The algebra of polynomials, from the quadratic on the SL paper to the higher-degree factorisations that decide HL Paper 2 questions. We move through the three forms of a quadratic, read off the shape and roots from the discriminant, exploit Vieta's relations to sidestep the quadratic formula, solve inequalities by sign analysis, and finally lift those tools to cubics and beyond via the factor theorem, the remainder theorem, and polynomial division.本单元讲多项式的代数：从 SL 试卷的二次函数，到决定 HL Paper 2 题目的高次因式分解。我们依次讲解二次函数三种形式、由判别式判断图像与根、用 Vieta 关系绕开求根公式、用符号分析解不等式，最后用因式定理、余数定理与多项式除法把这些工具推广到三次乃至更高次的情形。
How to use this guide本指南使用说明
B2 is the algebraic backbone of the Functions block. Quadratics appear on every Paper 1 and every Paper 2, and the HL extension to cubics and quartics shows up in algebraic and geometric problems alike. Each section pairs a cram-mode summary with one or two worked examples; the going-deeper blocks and pitfalls are flagged for the 7 chaser.B2 是函数模块的代数支柱。二次函数出现在每一份 Paper 1 与 Paper 2 上，HL 延伸到三次与四次的内容也在代数题与几何题中频繁登场。每节都配 cram 模式摘要加 1 到 2 道例题；"深入"小节与陷阱提示是给冲 7 分的同学准备的。
Memorise the three forms of a quadratic. Memorise the discriminant $\Delta = b^{2} - 4ac$ and what each sign tells you. Memorise Vieta's two relations $\alpha + \beta = -b/a$ and $\alpha\beta = c/a$. Practise one inequality by the "find roots, sketch, read off" routine.
背熟二次函数三种形式。背熟判别式 $\Delta = b^{2} - 4ac$ 及其符号含义。背熟 Vieta 两关系 $\alpha + \beta = -b/a$、$\alpha\beta = c/a$。按"求根、画图、读区间"做一道不等式练习。
Master completing the square as a one-step move; it is the same trick that proves the quadratic formula. Use Vieta to compute symmetric expressions in the roots without solving for the roots explicitly. For HL, internalise the equivalence between "$(x - r)$ is a factor" and "$P(r) = 0$", and use it to crack cubics on Paper 2.
把配方法练成一步到位的动作；它正是证明求根公式的同一招。用 Vieta 不解出根也能算根的对称表达式。HL 部分须把"$(x - r)$ 是因式"与"$P(r) = 0$"完全等同，并以此攻克 Paper 2 上的三次方程题。
Quadratic Functions: Three Forms二次函数的三种形式 SL 2.6
Conversion route. Standard to vertex is completing the square. Standard to factored is using the quadratic formula (or factoring by inspection). Vertex to standard is expanding the binomial.
转换路径。标准式转顶点式：配方法。标准式转因式式：求根公式（或观察分解）。顶点式转标准式：展开二次式。
Express $y = 2 x^{2} - 8 x + 5$ in vertex form and state the coordinates of the vertex.把 $y = 2 x^{2} - 8 x + 5$ 化为顶点式，并写出顶点坐标。
Step 1. Factor the leading coefficient from the $x$ terms.
第 1 步：从含 $x$ 的项中提取首项系数。
$$ y \;=\; 2 (x^{2} - 4 x) + 5. $$Step 2. Complete the square inside the bracket. Half of $-4$ is $-2$, squared is $4$.
第 2 步：在括号内配方。$-4$ 的一半为 $-2$，平方为 $4$。
$$ y \;=\; 2 \bigl[(x - 2)^{2} - 4\bigr] + 5 \;=\; 2 (x - 2)^{2} - 8 + 5 \;=\; 2 (x - 2)^{2} - 3. $$Result. Vertex form is $y = 2 (x - 2)^{2} - 3$. The vertex is at $(2, -3)$, and because $a = 2 > 0$ the parabola opens upward, so $(2, -3)$ is a minimum.
结果。顶点式为 $y = 2 (x - 2)^{2} - 3$。顶点 $(2, -3)$；因 $a = 2 > 0$，开口向上，故 $(2, -3)$ 为最小值点。
Going deeper: completing the square in general深入：一般情形下的配方法
For a general quadratic $y = a x^{2} + b x + c$ with $a \ne 0$, factor $a$ from the first two terms and add then subtract $(b/(2a))^{2}$ inside:
对一般二次式 $y = a x^{2} + b x + c$（$a \ne 0$），从前两项提 $a$，并在括号内加减 $(b/(2a))^{2}$：
$$ y \;=\; a \left( x + \tfrac{b}{2a} \right)^{2} + c - \tfrac{b^{2}}{4a}. $$Hence the vertex sits at $\bigl( -\tfrac{b}{2a}, \; c - \tfrac{b^{2}}{4a} \bigr)$. Notice that the $x$-coordinate of the vertex is exactly $-b/(2a)$, which is the axis of symmetry. The $y$-coordinate equals $-\Delta / (4a)$ where $\Delta = b^{2} - 4 a c$. This single calculation is the source of every shortcut later in the unit.
故顶点坐标为 $\bigl( -\tfrac{b}{2a}, \; c - \tfrac{b^{2}}{4a} \bigr)$。注意顶点横坐标 $-b/(2a)$ 正是对称轴方程。纵坐标等于 $-\Delta / (4a)$，其中 $\Delta = b^{2} - 4 a c$。本单元之后的所有捷径都源于这一次配方。
Discriminant and Quadratic Formula判别式与求根公式 SL 2.7
- $\Delta > 0$: two distinct real roots. The parabola cuts the $x$-axis twice.
- $\Delta = 0$: one repeated real root. The parabola touches the $x$-axis (tangent).
- $\Delta < 0$: no real roots. The parabola sits entirely above (if $a > 0$) or below (if $a < 0$) the $x$-axis.
- $\Delta > 0$：两不等实根。抛物线交 $x$ 轴两次。
- $\Delta = 0$：一重根。抛物线与 $x$ 轴相切。
- $\Delta < 0$：无实根。抛物线整体在 $x$ 轴上方（若 $a > 0$）或下方（若 $a < 0$）。
Find the values of $k$ for which $x^{2} + (k - 2) x + 9 = 0$ has a repeated real root.求 $k$，使 $x^{2} + (k - 2) x + 9 = 0$ 有重根。
Apply the discriminant condition. Repeated root requires $\Delta = 0$. Here $a = 1$, $b = k - 2$, $c = 9$.
套用判别式条件。有重根需 $\Delta = 0$。此处 $a = 1$、$b = k - 2$、$c = 9$。
$$ \Delta \;=\; (k - 2)^{2} - 4 \cdot 1 \cdot 9 \;=\; (k - 2)^{2} - 36 \;=\; 0. $$Solve for $k$.
解出 $k$。
$$ (k - 2)^{2} \;=\; 36 \quad\Longrightarrow\quad k - 2 \;=\; \pm 6 \quad\Longrightarrow\quad k \;=\; 8 \;\; \text{or} \;\; k \;=\; -4. $$Result. The equation has a repeated root when $k = 8$ or $k = -4$.
结果。$k = 8$ 或 $k = -4$ 时方程有重根。
Sum and Product of Roots (Vieta)根的和与积（Vieta 定理） SL 2.7
Symmetric expressions. Any symmetric polynomial in $\alpha$ and $\beta$ can be written using $\alpha + \beta$ and $\alpha\beta$ alone. Two identities show up constantly: $$ \alpha^{2} + \beta^{2} \;=\; (\alpha + \beta)^{2} - 2\alpha\beta, \qquad \frac{1}{\alpha} + \frac{1}{\beta} \;=\; \frac{\alpha + \beta}{\alpha\beta}. $$ Reverse Vieta. Given sum $S$ and product $P$, the quadratic with those roots is $x^{2} - S x + P = 0$.
对称表达式。$\alpha, \beta$ 的任何对称多项式都可用 $\alpha + \beta$ 与 $\alpha\beta$ 表出。两条恒等式最常用： $$ \alpha^{2} + \beta^{2} \;=\; (\alpha + \beta)^{2} - 2\alpha\beta, \qquad \frac{1}{\alpha} + \frac{1}{\beta} \;=\; \frac{\alpha + \beta}{\alpha\beta}. $$ 反向使用 Vieta。已知和 $S$ 与积 $P$，以 $\alpha, \beta$ 为根的二次方程为 $x^{2} - S x + P = 0$。
Let $\alpha, \beta$ be the roots of $x^{2} - 5 x + 6 = 0$. Find the value of $\alpha^{2} + \beta^{2}$ without solving for the roots.设 $\alpha, \beta$ 是 $x^{2} - 5 x + 6 = 0$ 的根。不解出根，求 $\alpha^{2} + \beta^{2}$。
Read off the sum and product. With $a = 1$, $b = -5$, $c = 6$:
读出和与积。$a = 1$、$b = -5$、$c = 6$：
$$ \alpha + \beta \;=\; -\frac{-5}{1} \;=\; 5, \qquad \alpha \beta \;=\; \frac{6}{1} \;=\; 6. $$Apply the symmetric identity.
套用对称恒等式。
$$ \alpha^{2} + \beta^{2} \;=\; (\alpha + \beta)^{2} - 2 \alpha \beta \;=\; 5^{2} - 2 \cdot 6 \;=\; 25 - 12 \;=\; 13. $$Cross-check. Solving gives $\alpha = 2$, $\beta = 3$ and $4 + 9 = 13$, which matches.
互相检验。实解 $\alpha = 2$、$\beta = 3$，$4 + 9 = 13$，吻合。
Going deeper: why Vieta works深入：Vieta 关系的来源
Divide the quadratic by $a$ to put it in monic form, then write it in factored form using its two roots:
将二次方程除以 $a$ 化为首一形式，再用两根写成因式形式：
$$ x^{2} + \tfrac{b}{a} x + \tfrac{c}{a} \;=\; (x - \alpha)(x - \beta) \;=\; x^{2} - (\alpha + \beta) x + \alpha \beta. $$Comparing coefficients of $x$ gives $\alpha + \beta = -b/a$, and comparing constant terms gives $\alpha\beta = c/a$. Hence Vieta's relations are just a coefficient match between standard and factored forms.
比较 $x$ 的系数得 $\alpha + \beta = -b/a$，比较常数项得 $\alpha\beta = c/a$。Vieta 关系即标准式与因式式之间的系数对应。
Quadratic Inequalities二次不等式 SL 2.7
- Move everything to one side so the inequality reads "quadratic vs $0$".
- Find the roots by factoring or by the quadratic formula. Let them be $r_{1} < r_{2}$ (assume $\Delta > 0$ for now).
- Sketch the parabola and read off the intervals where the inequality holds.
- Negative between the roots: $a x^{2} + b x + c < 0 \iff r_{1} < x < r_{2}$.
- Positive outside the roots: $a x^{2} + b x + c > 0 \iff x < r_{1}$ or $x > r_{2}$.
- 移项使不等式形如"二次 vs $0$"。
- 求根：因式分解或用求根公式。设 $r_{1} < r_{2}$（暂设 $\Delta > 0$）。
- 画抛物线，读取满足不等式的区间。
- 两根之间为负：$a x^{2} + b x + c < 0 \iff r_{1} < x < r_{2}$。
- 两根之外为正：$a x^{2} + b x + c > 0 \iff x < r_{1}$ 或 $x > r_{2}$。
Solve the inequality $x^{2} - x - 6 < 0$, giving your answer as an interval.解不等式 $x^{2} - x - 6 < 0$，用区间形式给出答案。
Factor. $x^{2} - x - 6 = (x - 3)(x + 2)$. Roots are $r_{1} = -2$ and $r_{2} = 3$.
因式分解。$x^{2} - x - 6 = (x - 3)(x + 2)$。根为 $r_{1} = -2$、$r_{2} = 3$。
Sketch. The leading coefficient is $a = 1 > 0$, so the parabola opens upward; it is below the $x$-axis precisely between the roots.
画图。首项系数 $a = 1 > 0$，开口向上；在两根之间正好低于 $x$ 轴。
Read off the interval.
读出区间。
$$ x^{2} - x - 6 < 0 \quad\Longleftrightarrow\quad -2 \;<\; x \;<\; 3. $$Result. The solution set is the open interval $(-2, 3)$.
结果。解集为开区间 $(-2, 3)$。
Factor and Remainder Theorems因式定理与余数定理 HL AHL 2.12
Factor theorem. $(x - r)$ is a factor of $P(x)$ if and only if $P(r) = 0$.
Translation. The factor theorem is the remainder theorem in the special case "remainder is zero". Together they convert root-finding into evaluation: to test whether $r$ is a root, just compute $P(r)$.
Use case. For a cubic $P(x) = x^{3} + \cdots$, try integer values $r \in \{ \pm 1, \pm 2, \pm 3, \ldots \}$ that divide the constant term. The first $r$ with $P(r) = 0$ gives a linear factor $(x - r)$; divide out to get a quadratic, then factor the quadratic by the methods of B2.1 to B2.4.
因式定理。$(x - r)$ 是 $P(x)$ 的因式当且仅当 $P(r) = 0$。
解读。因式定理是余数定理"余数为零"的特例。两者将"求根"转化为"求值"：检验 $r$ 是否为根，只需算 $P(r)$。
用法。对三次 $P(x) = x^{3} + \cdots$，试取能整除常数项的整数 $r \in \{ \pm 1, \pm 2, \pm 3, \ldots \}$。首个使 $P(r) = 0$ 的 $r$ 给出一次因式 $(x - r)$；做除法得二次，再用 B2.1 到 B2.4 的方法分解二次。
Show that $(x - 2)$ is a factor of $P(x) = x^{3} - 7 x + 6$, then factor $P(x)$ fully.证明 $(x - 2)$ 是 $P(x) = x^{3} - 7 x + 6$ 的因式，然后将 $P(x)$ 完全分解。
Apply the factor theorem. Compute $P(2)$:
用因式定理。算 $P(2)$：
$$ P(2) \;=\; 2^{3} - 7 \cdot 2 + 6 \;=\; 8 - 14 + 6 \;=\; 0. $$Since $P(2) = 0$, the factor theorem gives that $(x - 2)$ is a factor.
$P(2) = 0$，由因式定理，$(x - 2)$ 是因式。
Divide out the factor. Polynomial long division (or synthetic division) yields
除去该因式。多项式长除法（或综合除法）得
$$ x^{3} - 7 x + 6 \;=\; (x - 2)(x^{2} + 2 x - 3). $$Factor the quadratic. $x^{2} + 2 x - 3 = (x - 1)(x + 3)$.
分解二次。$x^{2} + 2 x - 3 = (x - 1)(x + 3)$。
Result.
结果。
$$ P(x) \;=\; (x - 2)(x - 1)(x + 3). $$The three roots are $x = 2$, $x = 1$, $x = -3$.
三根为 $x = 2$、$x = 1$、$x = -3$。
Going deeper: proof of the remainder theorem深入：余数定理的证明
Apply the division algorithm: dividing $P(x)$ by the linear divisor $(x - r)$ produces a quotient $Q(x)$ and a remainder $R$ of degree strictly less than $1$, hence a constant. Therefore
用带余除法：$P(x)$ 除以一次式 $(x - r)$ 得商 $Q(x)$ 与余数 $R$，余数的次数严格小于 $1$，故为常数。因此
$$ P(x) \;=\; (x - r) \, Q(x) + R. $$Substituting $x = r$ collapses the first term, leaving $P(r) = R$. Hence the remainder $R$ equals $P(r)$, proving the remainder theorem. The factor theorem is the corollary "$R = 0 \iff (x - r)$ divides $P(x)$".
代入 $x = r$，首项消失，得 $P(r) = R$。故余数 $R = P(r)$，余数定理得证。因式定理即推论"$R = 0 \iff (x - r) \mid P(x)$"。
Polynomial Division and Roots多项式除法与根 HL AHL 2.12
Finding the remaining roots. Once a known root $r$ has been pulled out, divide $P(x)$ by $(x - r)$ to obtain $Q(x)$ of one lower degree. Find the roots of $Q(x)$ by the methods of B2.1 to B2.4 (or by recursion if $Q(x)$ is still cubic or higher).
Polynomial inequalities by sign chart. Factor the polynomial completely. Mark all roots on a number line. The sign on each interval between adjacent roots is constant; determine it at a single test point per interval (or use the rule that a simple root flips the sign while a double root preserves it). Read off the solution.
求剩余的根。提出已知根 $r$ 后，将 $P(x)$ 除以 $(x - r)$，得到次数减一的 $Q(x)$。再用 B2.1 至 B2.4 的方法求 $Q(x)$ 的根（若 $Q(x)$ 仍为三次或更高次，则递归继续）。
多项式不等式（符号表法）。把多项式完全分解。在数轴上标出所有根。相邻根之间的区间内符号恒定；每个区间取一测试点确定符号（或用规则：单根处变号、二重根处不变号）。读出解集。
Solve the inequality $x^{3} - 7 x + 6 > 0$, using the factorisation $x^{3} - 7 x + 6 = (x - 2)(x - 1)(x + 3)$ obtained in B2.5.利用 B2.5 中得到的分解 $x^{3} - 7 x + 6 = (x - 2)(x - 1)(x + 3)$，解不等式 $x^{3} - 7 x + 6 > 0$。
List the roots in order. $-3 < 1 < 2$.
把根按从小到大排列。$-3 < 1 < 2$。
Sign chart. The cubic has positive leading coefficient, so its sign on the far right ($x > 2$) is positive. Each simple root flips the sign as we cross it from right to left.
符号表。三次的首项系数为正，故最右端（$x > 2$）符号为正。由右向左跨过每个单根时符号翻转。
| $x$ | $x < -3$ | $-3 < x < 1$ | $1 < x < 2$ | $x > 2$ |
|---|---|---|---|---|
| $P(x)$ | $-$ | $+$ | $-$ | $+$ |
Read off the solution. $P(x) > 0$ on the intervals where the sign is $+$:
读出解集。$P(x) > 0$ 对应符号为 $+$ 的区间：
$$ x^{3} - 7 x + 6 \;>\; 0 \quad\Longleftrightarrow\quad -3 \;<\; x \;<\; 1 \;\; \text{or} \;\; x \;>\; 2. $$Cross-check. Test $x = 0$: $0 - 0 + 6 = 6 > 0$, and $0$ lies in $(-3, 1)$. Test $x = 3$: $27 - 21 + 6 = 12 > 0$, and $3 > 2$. Both consistent.
互相检验。取 $x = 0$：$0 - 0 + 6 = 6 > 0$，且 $0 \in (-3, 1)$。取 $x = 3$：$27 - 21 + 6 = 12 > 0$，且 $3 > 2$。均一致。
Going deeper: polynomial long division step by step深入：多项式长除法的逐步执行
Divide $x^{3} - 7 x + 6$ by $(x - 2)$. Arrange the dividend with explicit zero placeholder for the missing $x^{2}$ term: $x^{3} + 0 \cdot x^{2} - 7 x + 6$.
用 $(x - 2)$ 除 $x^{3} - 7 x + 6$。把被除式补齐缺项：$x^{3} + 0 \cdot x^{2} - 7 x + 6$。
Iteration 1. $x^{3} \div x = x^{2}$. Multiply: $x^{2} (x - 2) = x^{3} - 2 x^{2}$. Subtract from the dividend: $(x^{3} + 0 \cdot x^{2}) - (x^{3} - 2 x^{2}) = 2 x^{2}$. Bring down the next term: $2 x^{2} - 7 x$.
第 1 次。$x^{3} \div x = x^{2}$。乘：$x^{2} (x - 2) = x^{3} - 2 x^{2}$。相减：$(x^{3} + 0 \cdot x^{2}) - (x^{3} - 2 x^{2}) = 2 x^{2}$。带下一项：$2 x^{2} - 7 x$。
Iteration 2. $2 x^{2} \div x = 2 x$. Multiply: $2 x (x - 2) = 2 x^{2} - 4 x$. Subtract: $(2 x^{2} - 7 x) - (2 x^{2} - 4 x) = -3 x$. Bring down: $-3 x + 6$.
第 2 次。$2 x^{2} \div x = 2 x$。乘：$2 x (x - 2) = 2 x^{2} - 4 x$。相减：$(2 x^{2} - 7 x) - (2 x^{2} - 4 x) = -3 x$。带下：$-3 x + 6$。
Iteration 3. $-3 x \div x = -3$. Multiply: $-3 (x - 2) = -3 x + 6$. Subtract: $(-3 x + 6) - (-3 x + 6) = 0$. Remainder is zero, confirming $(x - 2)$ is a factor.
第 3 次。$-3 x \div x = -3$。乘：$-3 (x - 2) = -3 x + 6$。相减：$(-3 x + 6) - (-3 x + 6) = 0$。余数为零，验证 $(x - 2)$ 是因式。
The quotient is $x^{2} + 2 x - 3$, matching the factorisation used in B2.5.
商为 $x^{2} + 2 x - 3$，与 B2.5 中所用的分解一致。
Exam Strategy and Common Pitfalls考试策略与常见陷阱
- Read the question for the form it wants. "Find the vertex" calls for vertex form; "find the roots" calls for factored form or the quadratic formula; "find $f(0)$" needs only the standard form constant term.
- 看题决定用哪种形式。"求顶点"用顶点式；"求根"用因式式或求根公式；"求 $f(0)$"只需看标准式的常数项。
- Use the discriminant for "find values of $k$". Conditions like "exactly one root", "two real roots", "no real roots" translate directly into $\Delta = 0$, $\Delta > 0$, $\Delta < 0$ on the parameter.
- "求 $k$ 的取值"题用判别式。"恰一根"、"两实根"、"无实根"分别对应 $\Delta = 0$、$\Delta > 0$、$\Delta < 0$。
- Vieta replaces the quadratic formula for symmetric problems. If the question asks for $\alpha^{2} + \beta^{2}$ or $1/\alpha + 1/\beta$, use Vieta. Solving for the roots first is slower and accumulates rounding error.
- 对称问题用 Vieta 替代求根公式。题目若问 $\alpha^{2} + \beta^{2}$ 或 $1/\alpha + 1/\beta$，用 Vieta。先解出根再代入更慢且累积舍入误差。
- Inequalities: factor first, sketch second, read off third. Algebraic shortcuts like "divide both sides by $x$" can lose roots or flip the inequality (when the divisor is negative). The factor-and-sketch routine never goes wrong.
- 不等式：先因式、再画图、再读区间。"两边除以 $x$"之类的代数捷径可能丢根或翻不等号（除式为负时）。"因式加画图"流程不会出错。
- Hunt for integer roots that divide the constant term. For a cubic $x^{3} + a x^{2} + b x + c$ with integer coefficients, any rational root $p/q$ (in lowest terms) has $p \mid c$ and $q \mid 1$. So test $\pm 1, \pm 2, \ldots$ among the divisors of $c$.
- 从整除常数项的整数中找根。整系数三次 $x^{3} + a x^{2} + b x + c$ 的任何有理根 $p/q$（最简形式）满足 $p \mid c$、$q \mid 1$。在 $c$ 的因数中试 $\pm 1, \pm 2, \ldots$ 即可。
- Polynomial inequalities use the sign chart, never algebraic cancellation. "Divide both sides by $(x - r)$" loses sign information when $(x - r)$ can be negative.
- 多项式不等式用符号表，绝不用代数消项。"两边除以 $(x - r)$"会在 $(x - r)$ 可为负时丢失符号信息。
- Show the long division if marks are allocated for working. A bare quotient with no division on the page risks losing the M1.
- 若评分含"过程分"，写出长除法。只给商而不写除法过程会丢 M1。
Flashcards闪卡
Unit B2 Practice Quiz单元 B2 练习测验
Readiness Checklist备考清单
Tick each item when you can do it cold, without notes, on your first attempt.
每一条都要"裸做"做对（不看笔记、一次过）才打勾。
- State the three forms of a quadratic and identify which feature each one exposes写出二次函数的三种形式并指出各自所突出的特征
- Convert any quadratic from standard form to vertex form by completing the square用配方法把任意二次从标准式化为顶点式
- Compute the discriminant and classify the number of real roots算判别式并分类实根数
- Use the discriminant to find values of a parameter $k$ that produce a specified root structure用判别式求参数 $k$ 使方程具有指定根的结构
- Apply the quadratic formula correctly, including with negative leading coefficient正确使用求根公式，含首项系数为负的情形
- Use Vieta's relations to compute $\alpha + \beta$, $\alpha\beta$, $\alpha^{2} + \beta^{2}$, $1/\alpha + 1/\beta$ without solving用 Vieta 关系不解根算 $\alpha + \beta$、$\alpha\beta$、$\alpha^{2} + \beta^{2}$、$1/\alpha + 1/\beta$
- Construct a quadratic whose roots have a given sum and product构造一个二次方程，使其两根有给定的和与积
- Solve a quadratic inequality by the factor-and-sketch routine用"因式加画图"流程解二次不等式
- Distinguish strict from non-strict inequalities and use the correct interval notation区分严格与非严格不等号，使用正确的区间记号
- HL State and apply the factor theorem: $(x - r)$ is a factor of $P(x)$ iff $P(r) = 0$陈述并应用因式定理：$(x - r) \mid P(x) \iff P(r) = 0$
- HL Apply the remainder theorem to compute the remainder on division by a linear factor用余数定理算除以一次因式的余数
- HL Perform polynomial long division, including with placeholder zeros for missing terms做多项式长除法，含缺项需补 $0$ 的情形
- HL Solve a cubic inequality by factoring fully and reading a sign chart完全分解三次后用符号表解三次不等式
IB Paper-Style PracticeIB 试卷风格练习
B2 Practice and Solutions are on the roadmap. They will ship under Practice Questions/Unit_B2_*.html with the bilingual built-in pattern.
B2 配套的 Practice 与 Solutions 已在排期，上线后位于 Practice Questions/Unit_B2_*.html，采用双语内嵌格式。