High School Math

Polynomial
Functions多项式函数

Polynomials are the natural generalisation of the parabola you mastered in the Quadratic Functions guide. This unit names the parts (degree, leading coefficient, constant term), does the arithmetic (addition, multiplication, long and synthetic division), and then unlocks roots through the remainder theorem, the factor theorem, and the rational root theorem. From there we graph by zeros, multiplicity, and end behaviour, count turning points against the degree, and finish with complex conjugate roots and polynomial inequalities via sign analysis. The unit feeds into Rational and Radical Expressions, where polynomial factoring rebuilds the numerator and denominator, AP Calc Unit 2 (derivative of a polynomial), and IB Math AA HL B2 (polynomial functions with the full factor / remainder / sum-and-product machinery).多项式（polynomial）是你在二次函数单元掌握的抛物线的自然推广。本单元先命名各部分：次数（degree）、首项系数（leading coefficient）、常数项（constant term），再做运算（加、乘、长除法与综合除法），然后用余数定理（remainder theorem）、因式定理（factor theorem）和有理根定理（rational root theorem）打开求根。接下来按零点、重根（multiplicity）与末端行为（end behaviour）作图，按次数估拐点（turning point）数量，最后以复共轭根（complex conjugate root）和符号分析下的多项式不等式（polynomial inequality）收尾。本单元承接的下游课程包括：有理与根式表达式（多项式因式分解重建分子与分母）、AP Calculus Unit 2（多项式求导）和 IB Math AA HL B2（多项式函数，完整的因式 / 余数 / 根与系数机制）。

US Common Core · ON · BC · ABUS 共同核心 · ON · BC · AB 7 sections · honors block on inequalities + complex roots7 节内容 · 不等式与复根为荣誉级
Read me first先读这里

How to use this guide如何使用本指南

Polynomial functions span Grades 11 and 12 in the curricula we map to: the factoring infrastructure sits in BC PC11 and US Algebra 2, but the named theorems (factor theorem, remainder theorem) live in BC PC12 and Ontario MHF4U, with the rational root theorem usually appearing in US Pre-Calc / honors. The table below tells you which sections of this guide are on your syllabus right now. Each row cites the curriculum document we checked it against, so the recommendation is grounded, not guessed.多项式函数横跨我们对照的大纲的 11、12 年级：因式分解基础放在 BC PC11 与美国 Algebra 2；而具名定理（因式定理、余数定理）位于 BC PC12 与安大略 MHF4U；有理根定理通常出现在美国 Pre-Calc / 荣誉课程中。下表告诉你本指南当前大纲下应学的章节。每行都列出我们对照的课纲文件，使建议有据可依。

If you are in…如果你在… Focus on these sections重点学习 Defer / skip可推迟 Source依据
🇨🇦 ON Grade 11 , MCR3U安大略 11 年级 , MCR3U (light polynomial exposure)（轻量多项式接触） §1 (vocabulary), §2 (arithmetic). Treat as a vocabulary primer so MHF4U's polynomial strand lands cleanly next year.§1（术语）、§2（运算）。作为术语预习，下一年 MHF4U 的多项式单元就能顺接。 §3-§7 are MHF4U work. Come back after MCR3U closes.§3-§7 属 MHF4U 内容。MCR3U 收尾后再回来。 math_grades_11-12_extract.md , MCR3U A1.1, A1.2 (function-notation foundation); polynomial work flagged for MHF4U, MCR3U A1.1、A1.2（函数记号基础）；多项式工作归 MHF4U
🇨🇦 ON Grade 12 , MHF4U安大略 12 年级 , MHF4U (deep)（深入） Full §1-§7. MHF4U's dedicated strand is "Polynomial and Rational Functions" and includes the factor theorem, the remainder theorem, characteristics of polynomial graphs, and polynomial inequalities.完整 §1-§7。MHF4U 的专属单元 "多项式与有理函数" 包括因式定理、余数定理、多项式图象特征以及多项式不等式。 Nothing , every section maps to an MHF4U expectation in the polynomial strand.无 , 每节都对应 MHF4U 多项式单元中的期望。 math_grades_11-12_extract.md , MHF4U strand Polynomial and Rational Functions (verbatim strand name in the extract), MHF4U 单元 多项式与有理函数（摘录原文单元名）
🇨🇦 BC Grade 12 , Pre-Calc 12BC 12 年级 , Pre-Calc 12 (deep)（深入） Full §1-§7. BC PC12 is the dedicated polynomial home; the curriculum lists the factor theorem and remainder theorem by name, and explicitly enumerates degree / extrema / zeros / end-behaviour as the graph characteristics.完整 §1-§7。BC PC12 是多项式的专属归属；课纲点名因式定理与余数定理，并将次数 / 极值 / 零点 / 末端行为列为图象特征。 Nothing , every section maps to a PC12 Content elaboration verbatim.无 , 每节都原文对应 PC12 内容阐释。 pc12_elab_extract.md , Content polynomial functions and equations (factor theorem, remainder theorem, degree, extrema, zeros, end-behaviour, algebraic and graphical solving), 内容 多项式函数与方程（因式定理、余数定理、次数、极值、零点、末端行为、代数与图象求解）
🇺🇸 US Algebra 2 (Grade 10-11)美国 Algebra 2（10-11 年级） (basic polynomial arithmetic)（基础多项式运算） §1 (vocabulary), §2 (arithmetic and long division), §3 (remainder + factor theorems are typical Alg 2 content), light §5 (graphing by zeros). §4 rational root theorem is Alg 2 in some scope-and-sequences, Pre-Calc in others.§1（术语）、§2（运算与长除法）、§3（余数与因式定理是典型 Alg 2 内容）、轻量 §5（按零点作图）。§4 有理根定理在部分大纲是 Alg 2 内容，其他归 Pre-Calc。 §6 complex conjugate roots is honors / Alg 2+; §7 polynomial inequalities are Pre-Calc in most US tracks.§6 复共轭根属 honors / Alg 2+；§7 多项式不等式在多数美国课程里归 Pre-Calc。 ccssm_hs_math_extract.md , HSA-SSE.A.1, HSF-IF.B.4, HSF-IF.C.8, HSA-REI.D.11 (polynomial graphical solving named verbatim), HSA-SSE.A.1、HSF-IF.B.4、HSF-IF.C.8、HSA-REI.D.11（多项式图象求解原文点名）
🇺🇸 US Pre-Calc / Honors美国 Pre-Calc / 荣誉 (factor + remainder + complex roots)（因式 + 余数 + 复根） Full §1-§7. Pre-Calc / Honors is the US course where the factor theorem, remainder theorem, rational root theorem, complex conjugate roots, and polynomial inequalities all land.完整 §1-§7。Pre-Calc / 荣誉是美国课程中安置因式定理、余数定理、有理根定理、复共轭根与多项式不等式的归属。 Nothing , treat this as the canonical US polynomial-functions unit.无 , 视为美国课程中标准的多项式函数单元。 ccssm_hs_math_extract.md , HSF-IF.C.7, HSF-IF.C.8, HSA-SSE.A.2, HSF-LE.A.3, HSF-BF.B.3, HSF-IF.C.7、HSF-IF.C.8、HSA-SSE.A.2、HSF-LE.A.3、HSF-BF.B.3
🇺🇸 US AP-feeder美国 AP 衔接 Full §1-§7. The end-behaviour / zeros / multiplicity vocabulary from §5 is exactly the AP Pre-Calculus framework; the polynomial-as-product-of-linear-and-quadratic-factors statement from §6 is what AP Calc Unit 2 takes for granted when differentiating polynomials.完整 §1-§7。§5 的末端行为 / 零点 / 重根术语正是 AP Pre-Calculus 的框架；§6 中"多项式为线性与二次因式之积"的论断，是 AP Calc Unit 2 在求多项式导数时默认成立的前提。 Nothing , cross-reference to AP Calc Unit 2 (derivative of a polynomial) and AP Pre-Calc Unit 1.无 , 交叉参考 AP Calc Unit 2（多项式求导）与 AP Pre-Calc Unit 1。 See What this feeds into; cross-link to AP Calculus/Study Guides/Unit_2_*见 后续单元；交叉链接到 AP Calculus/Study Guides/Unit_2_*
🇺🇸 SAT bound备考 SAT §1-§3 are the SAT-frequent slice: polynomial vocabulary, polynomial arithmetic, and the factor-theorem-as-"if-$p(a)-equals-zero-then-$x-minus-$a$-is-a-factor" question style. §5 graphing-by-zeros is SAT-common too.§1-§3 是 SAT 高频片段：多项式术语、多项式运算，以及"若 $p(a) = 0$ 则 $(x - a)$ 是因式"形式的因式定理题。§5 按零点作图在 SAT 也常见。 §6 (complex roots) and §7 (inequalities) are not on the SAT in this form.§6（复根）与 §7（不等式）以此形式不出现在 SAT。 See What this feeds into at the end of this unit for the SAT and AP cross-referencesSAT 与 AP 交叉参考见本单元末尾的 后续单元
IB / AP feederIB / AP 衔接 Full §1-§7 with depth. The factor theorem / remainder theorem / sum-and-product-of-roots block in §3 and §6 is exactly what IB Math HL B2 takes for granted; the binomial expansion connection to combinatorics is the IB HL A3 cross-reference.完整 §1-§7 并加深。§3 与 §6 的因式定理 / 余数定理 / 根与系数模块，正是 IB Math HL B2 所默认掌握的；二项式展开与组合学的连接是 IB HL A3 的交叉参考。 Nothing , this is the prerequisite block for IB HL B2 (Polynomial Functions) and AP Calc Unit 2.无 , 本单元是 IB HL B2（多项式函数）与 AP Calc Unit 2 的先修模块。 See feeders; targets include IB HL B2, IB HL A3, AP Calc U2.见后续单元；目标含 IB HL B2、IB HL A3、AP Calc U2。

Once you have located your row, use the two cards below for the speed at which you should work through the recommended sections.找到所在行后，用下面两张卡片决定推进速度。

!
If you are cramming the night before如果你在临阵磨枪

Memorise four facts. (1) The factor theorem: $p(a) = 0$ if and only if $(x - a)$ is a factor of $p(x)$. (2) The remainder theorem: the remainder when $p(x)$ is divided by $(x - a)$ is $p(a)$. (3) End behaviour is set by the leading term: $a_{n} x^{n}$ alone decides what happens as $x \to \pm \infty$. (4) A polynomial of degree $n$ has at most $n - 1$ turning points and exactly $n$ roots counted with multiplicity (in $\mathbb{C}$). Skim the cram-cheat boxes at the top of each section in your row.背熟四件事。（1）因式定理：$p(a) = 0$ 当且仅当 $(x - a)$ 是 $p(x)$ 的因式。（2）余数定理：$p(x)$ 除以 $(x - a)$ 的余数即 $p(a)$。（3）末端行为由首项决定：当 $x \to \pm \infty$ 时仅 $a_{n} x^{n}$ 起作用。（4）次数为 $n$ 的多项式至多有 $n - 1$ 个拐点，并在 $\mathbb{C}$ 中按重数恰有 $n$ 个根。逐节速读你所在行各节顶部的速记框。

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If you are going for the top mark如果你目标顶分

Always state the theorem you are invoking by name: "by the factor theorem…", "by the rational root theorem the candidates are…". Always check multiplicity when sketching: an even-multiplicity zero touches the axis without crossing; an odd-multiplicity zero crosses. Always state the leading-term end-behaviour signs as a one-line conclusion. Sign analysis (§7) transfers verbatim to rational-function and derivative-of-polynomial work.引用定理务必点名：“由因式定理…”、“由有理根定理，候选为…”。作图时务必检查重数：偶重数零点触轴不穿越；奇重数零点穿越。务必把首项末端行为的符号写成一行结论。§7 的符号分析可原样迁移到有理函数与多项式求导工作。

Honors flag.荣誉级标记。 Section 6 (complex roots and the complex conjugate root theorem) carries the Honors chip for US Algebra 2 / Pre-Calc / BC PC12 tracks. US Common Core treats complex numbers under HSN-CN; the polynomial-specific consequence ("non-real roots of a real polynomial come in conjugate pairs") sits above the linear-scoped CCSSM extract but is standard in US Pre-Calc and is named directly in BC PC12 via the polynomial-equations elaboration. Section 7 (polynomial inequalities) is an Honors extension for US Algebra 2 (BC PC11's sign-analysis elaboration generalises naturally; the PC12 extract names polynomial equation solving algebraically and graphically without an explicit inequality bullet, so this is the PC11 sign-analysis technique applied at higher degree).第 6 节（复根与复共轭根定理）在美国 Algebra 2 / Pre-Calc / BC PC12 课程中标为 Honors。美国共同核心把复数放在 HSN-CN；针对多项式的推论（"实多项式的非实根成共轭对")在线性范围的 CCSSM 摘录之上，但在美国 Pre-Calc 中是标准内容，且在 BC PC12 多项式方程阐释中直接点名。第 7 节（多项式不等式）对美国 Algebra 2 属 Honors 拓展（BC PC11 的符号分析阐释自然推广；PC12 摘录只点名代数与图象求解，未单列不等式条款，因此本节是把 PC11 符号分析技术应用到更高次）。
Section 1 · Vocabulary第 1 节 · 术语

Polynomial Vocabulary多项式术语 🇺🇸 CCSSM · 🇨🇦 ON MHF4U · 🇨🇦 BC PC12

A polynomial in $x$.$x$ 的多项式。 Finite sum of $a_{k} x^{k}$ with $k$ a non-negative integer:$a_{k} x^{k}$（$k$ 为非负整数）的有限和： $$ p(x) \;=\; a_{n} x^{n} + a_{n-1} x^{n-1} + \cdots + a_{1} x + a_{0}, \qquad a_{n} \ne 0. $$
  • Degree次数 $n$ , largest exponent with non-zero coefficient (linear $1$, quadratic $2$, cubic $3$, quartic $4$, quintic $5$)., 非零系数对应的最高次（一次 $1$、二次 $2$、三次 $3$、四次 $4$、五次 $5$）。
  • Leading coefficient首项系数 $a_{n}$ , controls end behaviour., 决定末端行为。
  • Constant term常数项 $a_{0} = p(0)$ , the $y$-intercept., 即 $y$ 截距。
  • Monic首一 , leading coefficient $1$. Terms vs. factors: terms are added, factors are multiplied., 首项系数为 $1$。项与因式：项是相加的，因式是相乘的。
End behaviour.末端行为。 Only $a_{n} x^{n}$ matters as $x \to \pm \infty$. Four cases by parity of $n$ and sign of $a_{n}$: even-positive both ends to $+\infty$ (like $x^{2}$); even-negative both to $-\infty$; odd-positive left $\to -\infty$, right $\to +\infty$ (like $x^{3}$); odd-negative the reverse.当 $x \to \pm \infty$ 时只看 $a_{n} x^{n}$。按 $n$ 的奇偶和 $a_{n}$ 的符号分四种：偶 + 正两端均趋 $+\infty$（似 $x^{2}$）；偶 + 负两端均趋 $-\infty$；奇 + 正左 $\to -\infty$、右 $\to +\infty$（似 $x^{3}$）；奇 + 负相反。
Syllabus note.大纲提示。 BC Pre-Calc 12 enumerates the named graph characteristics , degree, extrema, zeros, end-behaviour , verbatim in its polynomial-functions Content elaboration. That four-word list is the right vocabulary checklist for any BC PC12 polynomial question; the same vocabulary surfaces in MHF4U (Ontario) and in US Pre-Calc / AP Pre-Calc, often word-for-word.BC PC12 在多项式函数内容阐释中原文列出四项图象特征 , 次数、极值、零点、末端行为。这四个词是 BC PC12 任何多项式题的术语清单；同一组术语也常原文出现在 MHF4U（安大略）以及美国 Pre-Calc / AP Pre-Calc 中。
Worked Example 1 · Identify the parts例题 1 · 辨识各部分

For $p(x) = -3 x^{4} + 5 x^{2} - 2 x + 7$, state the degree, leading coefficient, constant term, and end behaviour.对 $p(x) = -3 x^{4} + 5 x^{2} - 2 x + 7$，写出次数、首项系数、常数项及末端行为。

Identify.辨识。 $p(x) = -3 x^{4} + 0 \cdot x^{3} + 5 x^{2} - 2 x + 7$ (note the missing $x^{3}$ term).$p(x) = -3 x^{4} + 0 \cdot x^{3} + 5 x^{2} - 2 x + 7$（注意 $x^{3}$ 项缺失）。

Degree次数 $= 4$. Leading coefficient首项系数 $a_{4} = -3$. Constant term常数项 $a_{0} = 7 = p(0)$.

End behaviour.末端行为。 $n = 4$ even and $a_{n} = -3 < 0$: both ends fall, $p(x) \to -\infty$ as $x \to \pm \infty$.$n = 4$ 为偶，$a_{n} = -3 < 0$：两端均下降，$x \to \pm \infty$ 时 $p(x) \to -\infty$。

Evaluate.小结。 Quartic, opens "down on both ends," $y$-intercept $(0, 7)$, at most $3$ turning points (general: degree $n \Rightarrow$ at most $n - 1$ turning points). The next sections find where it crosses zero.四次多项式，"两端向下"，$y$ 截距 $(0, 7)$，至多 $3$ 个拐点（一般规律：次数 $n \Rightarrow$ 至多 $n - 1$ 个拐点）。后续章节将求其零点。

Going deeper · Why only the leading term decides end behaviour深入 · 为何只有首项决定末端行为

Factor $a_{n} x^{n}$ out: $p(x) = a_{n} x^{n} \left( 1 + \frac{a_{n-1}}{a_{n} x} + \cdots + \frac{a_{0}}{a_{n} x^{n}} \right)$. As $|x| \to \infty$, every term in the bracket except the constant $1$ heads to $0$, so the bracket approaches $1$ and $p(x)$ behaves like $a_{n} x^{n}$ for large $|x|$. That is the whole "only the leading term matters" rule: lower-order terms are eventually negligible. The same argument explains why polynomials of different degree must diverge from each other , the algebraic content behind HSF-LE.A.3's "exponential eventually beats polynomial" claim, applied within the polynomial family itself.把 $a_{n} x^{n}$ 提出：$p(x) = a_{n} x^{n} \left( 1 + \frac{a_{n-1}}{a_{n} x} + \cdots + \frac{a_{0}}{a_{n} x^{n}} \right)$。当 $|x| \to \infty$ 时，括号内除常数 $1$ 外每一项均趋 $0$，故括号趋 $1$，$|x|$ 大时 $p(x)$ 表现如 $a_{n} x^{n}$。这就是"只看首项"的全部含义：低次项最终可忽略。同样的论证解释了不同次数的多项式必然彼此发散 , 这就是 HSF-LE.A.3"指数终将超越多项式"在多项式族内部的代数版本。

What is the end behaviour of $p(x) = 2 x^{5} - 7 x^{3} + x - 4$?$p(x) = 2 x^{5} - 7 x^{3} + x - 4$ 的末端行为是？
§1 · Q1
As $x \to \pm \infty$, $p(x) \to +\infty$.$x \to \pm \infty$ 时 $p(x) \to +\infty$。
As $x \to -\infty$, $p(x) \to +\infty$; as $x \to +\infty$, $p(x) \to -\infty$.$x \to -\infty$ 时 $p(x) \to +\infty$；$x \to +\infty$ 时 $p(x) \to -\infty$。
As $x \to -\infty$, $p(x) \to -\infty$; as $x \to +\infty$, $p(x) \to +\infty$.$x \to -\infty$ 时 $p(x) \to -\infty$；$x \to +\infty$ 时 $p(x) \to +\infty$。
As $x \to \pm \infty$, $p(x) \to -\infty$.$x \to \pm \infty$ 时 $p(x) \to -\infty$。
Degree $n = 5$ is odd, leading coefficient $a_{n} = 2 > 0$. Odd-positive: left tail down, right tail up.次数 $n = 5$ 为奇，首项系数 $a_{n} = 2 > 0$。奇 + 正：左尾下降，右尾上升。
Look only at the leading term $2 x^{5}$. Odd degree means the two tails go opposite directions; positive leading coefficient means the right tail rises.只看首项 $2 x^{5}$。奇次意味着两端方向相反；正首项系数意味着右端上升。
A polynomial $p(x)$ has degree $6$. The maximum possible number of turning points is:多项式 $p(x)$ 次数为 $6$。拐点数的最大可能值是：
§1 · Q2
$6$
$5$
$7$
$3$
A polynomial of degree $n$ has at most $n - 1$ turning points (one for each zero of the degree-$(n-1)$ derivative). For $n = 6$, that bound is $5$.次数 $n$ 的多项式至多有 $n - 1$ 个拐点（对应 $n-1$ 次导数的零点）。$n = 6$ 时上界为 $5$。
The bound is $n - 1$, set by the degree of the derivative. For $n = 6$, that is $5$.上界为 $n - 1$，由导数次数决定。$n = 6$ 时为 $5$。
Section 2 · Arithmetic and division第 2 节 · 运算与除法

Polynomial Arithmetic and Division多项式运算与除法

Addition and multiplication.加法与乘法。 Add by combining like terms (same exponent). Multiply by the distributive law term-by-term; the degree of a product is the sum of the degrees.相加合并同类项（同指数）；按分配律逐项相乘；乘积的次数等于各因式次数之和。

Long division.长除法。 To divide $p(x)$ by $d(x)$:用 $d(x)$ 除 $p(x)$：
  1. Arrange both in descending order, including a $0$ coefficient for any missing power.两者均按降幂排列，缺项用 $0$ 系数补齐。
  2. Divide the leading term of $p$ by the leading term of $d$ to get the next term of the quotient.$p$ 的首项除以 $d$ 的首项，得到商的下一项。
  3. Multiply $d$ by that term and subtract from $p$.用该项乘 $d$，从 $p$ 中减去。
  4. Repeat with the new "remainder so far" until its degree is less than $\deg d$.对新的"当前余数"重复以上步骤，直到其次数小于 $\deg d$。
The result is $p(x) = d(x) \cdot q(x) + r(x)$ with $\deg r < \deg d$.最终 $p(x) = d(x) \cdot q(x) + r(x)$，且 $\deg r < \deg d$。

Synthetic division.综合除法。 A streamlined shortcut for dividing by a linear factor $(x - a)$:除以线性因式 $(x - a)$ 的速记法：
  1. Write the coefficients of $p$ in descending order (use $0$ for missing powers).按降幂写出 $p$ 的系数（缺项补 $0$）。
  2. Put $a$ on the left.把 $a$ 放在左侧。
  3. Bring down the first coefficient. Multiply by $a$ and add to the next coefficient. Repeat.把首系数下移；乘 $a$ 加到下一系数；如此循环。
  4. The final number is $p(a)$ , the remainder. The other numbers, in order, are the coefficients of the quotient (degree one less than $p$).最后一个数即 $p(a)$ , 余数；其余数字按序为商的系数（次数比 $p$ 低 $1$）。
Synthetic division is exactly long division by $(x - a)$, rewritten as a number-crunching procedure.综合除法本质上就是除以 $(x - a)$ 的长除法，写成纯数字流程而已。
Worked Example 2 · Long division of $p(x) = 2 x^{3} - 5 x^{2} + 3 x - 7$ by $d(x) = x - 2$例题 2 · 用 $d(x) = x - 2$ 长除 $p(x) = 2 x^{3} - 5 x^{2} + 3 x - 7$

Find the quotient $q(x)$ and the remainder $r$.求商 $q(x)$ 与余数 $r$。

Set up.列式。 Coefficients of $p$: $2, -5, 3, -7$.$p$ 的系数：$2, -5, 3, -7$。

Step 1.第 1 步。 $2 x^{3} \div x = 2 x^{2}$; multiply by $d$: $2 x^{3} - 4 x^{2}$; subtract: $-x^{2}$. Bring down: dividend $-x^{2} + 3 x - 7$.$2 x^{3} \div x = 2 x^{2}$；乘 $d$ 得 $2 x^{3} - 4 x^{2}$；相减得 $-x^{2}$。下移：被除式变为 $-x^{2} + 3 x - 7$。

Step 2.第 2 步。 $-x^{2} \div x = -x$; multiply: $-x^{2} + 2 x$; subtract: $x$. Bring down: dividend $x - 7$.$-x^{2} \div x = -x$；乘得 $-x^{2} + 2 x$；相减得 $x$。下移：被除式变为 $x - 7$。

Step 3.第 3 步。 $x \div x = 1$; multiply: $x - 2$; subtract: $-5$. Degree $0 < \deg d$, stop.$x \div x = 1$；乘得 $x - 2$；相减得 $-5$。次数 $0 < \deg d$，停止。

Read off.读出结果。 $q(x) = 2 x^{2} - x + 1$, $r = -5$, so $2 x^{3} - 5 x^{2} + 3 x - 7 = (x - 2)(2 x^{2} - x + 1) - 5$. Sanity check at $x = 2$: left $= 16 - 20 + 6 - 7 = -5$, matching $r$ , previewing the remainder theorem in §3.$q(x) = 2 x^{2} - x + 1$、$r = -5$，故 $2 x^{3} - 5 x^{2} + 3 x - 7 = (x - 2)(2 x^{2} - x + 1) - 5$。$x = 2$ 处核验：左 $= 16 - 20 + 6 - 7 = -5$，与 $r$ 一致 , 这就是 §3 余数定理的预演。

Going deeper · Synthetic division on the same problem深入 · 同一题的综合除法

Coefficients $2, -5, 3, -7$. Divisor $(x - 2)$, so $a = 2$. Bring down $2$; multiply by $a = 2$ and add: $-5 + 4 = -1$. Continue: $3 + (-2) = 1$; $-7 + 2 = -5$. Read off the quotient coefficients $2, -1, 1$ (one degree lower than $p$) and the remainder $-5$, matching long division exactly. Synthetic division is faster but only works for divisors of the form $(x - a)$; for higher-degree divisors, fall back to long division.系数 $2, -5, 3, -7$。除式 $(x - 2)$，故 $a = 2$。下移 $2$；乘 $a = 2$ 后相加：$-5 + 4 = -1$。继续：$3 + (-2) = 1$；$-7 + 2 = -5$。读出商的系数 $2, -1, 1$（次数比 $p$ 低 $1$）与余数 $-5$，与长除法完全一致。综合除法更快，但只适用于 $(x - a)$ 形式的除式；对高次除式需退回长除法。

What is the remainder when $p(x) = x^{3} + 2 x^{2} - x + 5$ is divided by $(x + 1)$?$p(x) = x^{3} + 2 x^{2} - x + 5$ 除以 $(x + 1)$ 的余数是？
§2 · Q1
$5$
$7$
$-7$
$3$
Divisor $(x + 1) = (x - (-1))$, so $a = -1$. Synthetic division on coefficients $1, 2, -1, 5$ with $a = -1$ gives a final value (remainder) of $7$. (Cross-check via the remainder theorem in §3: $p(-1) = -1 + 2 + 1 + 5 = 7$.)除式 $(x + 1) = (x - (-1))$，故 $a = -1$。对系数 $1, 2, -1, 5$ 用 $a = -1$ 做综合除法，最后一个数（余数）为 $7$。（用 §3 余数定理核验：$p(-1) = -1 + 2 + 1 + 5 = 7$。）
When dividing by $(x + 1)$, use synthetic division with $a = -1$. The last number you produce is the remainder.除以 $(x + 1)$ 时综合除法取 $a = -1$。最后产生的数即余数。
If $p(x) = x^{4} - 16$ is divided by $(x - 2)$, the quotient is:$p(x) = x^{4} - 16$ 除以 $(x - 2)$ 的商是：
§2 · Q2
$x^{3} + 2 x^{2} + 4 x + 8$ with remainder $16$.$x^{3} + 2 x^{2} + 4 x + 8$，余数 $16$。
$x^{3} - 2 x^{2} + 4 x - 8$ with remainder $0$.$x^{3} - 2 x^{2} + 4 x - 8$，余数 $0$。
$x^{3} + 2 x^{2} + 4 x + 8$ with remainder $0$.$x^{3} + 2 x^{2} + 4 x + 8$，余数 $0$。
$x^{3} - 8$ with remainder $0$.$x^{3} - 8$，余数 $0$。
Synthetic division on $1, 0, 0, 0, -16$ with $a = 2$: $1$, then $0 + 2 = 2$, then $0 + 4 = 4$, then $0 + 8 = 8$, then $-16 + 16 = 0$. Quotient $x^{3} + 2 x^{2} + 4 x + 8$, remainder $0$.对 $1, 0, 0, 0, -16$ 用 $a = 2$ 做综合除法：$1$、$0 + 2 = 2$、$0 + 4 = 4$、$0 + 8 = 8$、$-16 + 16 = 0$。商 $x^{3} + 2 x^{2} + 4 x + 8$，余数 $0$。
Use synthetic division with $a = 2$. Remember to insert $0$ coefficients for the missing $x^{3}, x^{2}, x^{1}$ powers.综合除法用 $a = 2$。注意为缺失的 $x^{3}, x^{2}, x^{1}$ 项补 $0$ 系数。
Section 3 · Remainder and Factor Theorems第 3 节 · 余数定理与因式定理

The Remainder and Factor Theorems余数定理与因式定理 🇨🇦 ON MHF4U · 🇨🇦 BC PC12 core

Remainder theorem.余数定理。 When $p(x)$ is divided by $(x - a)$, the remainder equals $p(a)$.$p(x)$ 除以 $(x - a)$ 的余数等于 $p(a)$。 $$ p(x) \;=\; (x - a) \cdot q(x) + p(a). $$ Factor theorem.因式定理。 $(x - a)$ is a factor of $p(x)$ if and only if $p(a) = 0$. (Special case of the remainder theorem with $p(a) = 0$.)$(x - a)$ 是 $p(x)$ 的因式当且仅当 $p(a) = 0$。（余数定理在 $p(a) = 0$ 时的特例。）

How to use them.使用方法。
  • Check whether $a$ is a root检验 $a$ 是否为根 without doing long division: compute $p(a)$ and see if it is zero.无需长除：算 $p(a)$ 看是否为零。
  • Factor a polynomial分解多项式 once you have spotted a root $a$: divide by $(x - a)$ (synthetic division is fastest) to get the quotient $q(x)$, then continue factoring $q(x)$.：一旦找到一个根 $a$，除以 $(x - a)$（综合除法最快）得商 $q(x)$，再继续分解 $q(x)$。
  • Find an unknown coefficient求未知系数 when you know a root: substitute and solve.：已知一根时，代入求解。
Syllabus note.大纲提示。 BC Pre-Calc 12 names the factor theorem and the remainder theorem verbatim in its polynomial-functions Content elaboration , they are not a derivation here but a named curriculum learning standard. Ontario MHF4U's "Polynomial and Rational Functions" strand carries the same two theorems. In US Common Core they sit one click above the linear-scoped extract under HSA-APR (Arithmetic with Polynomials and Rational Expressions); the cluster name and category exist, but the granular bullets are not in ccssm_hs_math_extract.md. The mathematical content below is identical across all three curricula.BC PC12 在多项式函数内容阐释中原文点名 因式定理与余数定理 , 此处不是推导而是具名课纲标准。安大略 MHF4U 的"多项式与有理函数"单元也同样收录这两个定理。在美国共同核心中，它们位于线性范围摘录之上、归入 HSA-APR（多项式与有理表达式的运算）；簇名与类别存在，但细分条目不在 ccssm_hs_math_extract.md 内。下面的数学内容在三套大纲中完全一致。
Worked Example 3 · Factor a cubic by spotting a root例题 3 · 通过试根分解三次多项式

Fully factor $p(x) = x^{3} - 6 x^{2} + 11 x - 6$ over the reals.在实数范围内完全分解 $p(x) = x^{3} - 6 x^{2} + 11 x - 6$。

Hunt for a root.试根。 Try $x = 1$: $p(1) = 1 - 6 + 11 - 6 = 0$. By the factor theorem, $(x - 1)$ is a factor.试 $x = 1$：$p(1) = 1 - 6 + 11 - 6 = 0$。由因式定理，$(x - 1)$ 是因式。

Divide.作除。 Synthetic division with $a = 1$ on $1, -6, 11, -6$ gives quotient $x^{2} - 5 x + 6$ and remainder $0$. Factor the quotient: $(x - 2)(x - 3)$ (sum $5$, product $6$).对 $1, -6, 11, -6$ 用 $a = 1$ 做综合除法，商为 $x^{2} - 5 x + 6$，余数 $0$。分解商：$(x - 2)(x - 3)$（和 $5$，积 $6$）。

Assemble.合并结论。 $p(x) = (x - 1)(x - 2)(x - 3)$. Roots $1, 2, 3$, all simple. Vieta check on the cubic: sum $1 + 2 + 3 = 6 = -a_{n-1}/a_{n}$; product $1 \cdot 2 \cdot 3 = 6 = (-1)^{n} a_{0}/a_{n} = -(-6)$. Confirmed.$p(x) = (x - 1)(x - 2)(x - 3)$。三个根 $1, 2, 3$ 均为单根。三次 Vieta 核验：根之和 $1 + 2 + 3 = 6 = -a_{n-1}/a_{n}$；根之积 $1 \cdot 2 \cdot 3 = 6 = (-1)^{n} a_{0}/a_{n} = -(-6)$。确认。

Going deeper · Why the remainder theorem is true深入 · 余数定理为何成立

Long division of $p(x)$ by $(x - a)$ leaves a remainder $r(x)$ with $\deg r < \deg(x - a) = 1$, so $r$ is a constant: $p(x) = (x - a) q(x) + r$. Substitute $x = a$: $p(a) = 0 \cdot q(a) + r = r$. So the remainder equals $p(a)$ , the remainder theorem in one substitution. The factor theorem is the special case $p(a) = 0$: then $p(x) = (x - a) q(x)$, exhibiting $(x - a)$ as a factor; the converse is immediate.$p(x)$ 长除 $(x - a)$ 得到余式 $r(x)$，且 $\deg r < \deg(x - a) = 1$，故 $r$ 是常数：$p(x) = (x - a) q(x) + r$。代入 $x = a$：$p(a) = 0 \cdot q(a) + r = r$。故余数等于 $p(a)$ , 一次代入即得余数定理。因式定理是 $p(a) = 0$ 的特例：此时 $p(x) = (x - a) q(x)$，$(x - a)$ 显式为因式；反向亦显然。

If $p(x) = 2 x^{3} - 3 x^{2} + k x + 1$ has $(x - 1)$ as a factor, what is $k$?若 $p(x) = 2 x^{3} - 3 x^{2} + k x + 1$ 以 $(x - 1)$ 为因式，则 $k$ 为？
§3 · Q1
$k = 0$
$k = 1$
$k = -1$
$k = 2$
By the factor theorem, $(x - 1)$ a factor means $p(1) = 0$. Compute: $p(1) = 2 - 3 + k + 1 = k$. Set $k = 0$.由因式定理，$(x - 1)$ 为因式即 $p(1) = 0$。算得：$p(1) = 2 - 3 + k + 1 = k$。令 $k = 0$。
Use the factor theorem: $(x - 1)$ a factor of $p(x)$ iff $p(1) = 0$. Substitute and solve for $k$.用因式定理：$(x - 1) \mid p(x) \iff p(1) = 0$。代入解 $k$。
The remainder when $p(x) = x^{4} - 3 x^{2} + 2 x - 5$ is divided by $(x - 2)$ is:$p(x) = x^{4} - 3 x^{2} + 2 x - 5$ 除以 $(x - 2)$ 的余数是：
§3 · Q2
$0$
$5$
$-3$
$3$
By the remainder theorem, the remainder is $p(2) = 16 - 12 + 4 - 5 = 3$.由余数定理，余数为 $p(2) = 16 - 12 + 4 - 5 = 3$。
The remainder theorem says the remainder is $p(a)$; here $a = 2$. Substitute directly , no long division required.余数定理告诉我们余数即 $p(a)$；此处 $a = 2$。直接代入 , 无需长除。
Section 4 · Finding roots第 4 节 · 求根

Finding Roots: Rational Root Theorem and Factoring Strategies求根：有理根定理与因式分解策略

Rational root theorem.有理根定理。 Let $p(x) = a_{n} x^{n} + \cdots + a_{1} x + a_{0}$ have integer coefficients with $a_{n} \ne 0$ and $a_{0} \ne 0$. If $p$ has a rational root $\frac{p}{q}$ in lowest terms (so $\gcd(p, q) = 1$), then设 $p(x) = a_{n} x^{n} + \cdots + a_{1} x + a_{0}$ 系数均为整数，且 $a_{n} \ne 0$、$a_{0} \ne 0$。若 $p$ 有最简形式的有理根 $\frac{p}{q}$（即 $\gcd(p, q) = 1$），则
  • $p$ divides $a_{0}$ (the constant term), and$p \mid a_{0}$（常数项），且
  • $q$ divides $a_{n}$ (the leading coefficient).$q \mid a_{n}$（首项系数）。
The candidate list is finite: every $\pm (\text{factor of } a_{0}) / (\text{factor of } a_{n})$ in lowest terms.候选有限：所有最简形式的 $\pm (a_{0} \text{的因数}) / (a_{n} \text{的因数})$。

Strategy for solving $p(x) = 0$.求解 $p(x) = 0$ 的策略。
  1. Pull a GCF first先提公因式 if one is staring at you (factors out a power of $x$ to expose any $x = 0$ roots).（若一眼可见）。提出 $x$ 的幂可显式暴露 $x = 0$ 的根。
  2. Try special factorings尝试特殊因式: difference of squares $a^{2} - b^{2} = (a - b)(a + b)$; sum/difference of cubes $a^{3} \pm b^{3} = (a \pm b)(a^{2} \mp a b + b^{2})$; grouping for four-term expressions.：平方差 $a^{2} - b^{2} = (a - b)(a + b)$；立方和 / 差 $a^{3} \pm b^{3} = (a \pm b)(a^{2} \mp a b + b^{2})$；四项分组。
  3. List rational candidates列出有理候选 with the rational root theorem. Test each with synthetic division (which delivers $p(a)$ and the quotient simultaneously).：用有理根定理。每个候选用综合除法测试（同时给出 $p(a)$ 与商）。
  4. Recurse.递归。 Each rational root drops the degree by $1$; eventually you reach a quadratic, which you solve by §2.3-§2.5 techniques.每找到一个有理根次数降 $1$；最终降到二次，用 §2.3-§2.5 的方法求解。
  5. Irrational roots无理根 come from the residual quadratic via the quadratic formula. Complex roots come in conjugate pairs (§6).从残余二次中用求根公式获得。复根成共轭对（§6）。
Syllabus note.大纲提示。 The rational root theorem is standard in US Pre-Calc / Algebra 2 Honors; it is the canonical tool for hand-factoring a cubic or quartic with integer coefficients. BC Pre-Calc 12 lists "solving equations algebraically and graphically" in the polynomial Content elaboration without naming the theorem , the theorem is the standard solving technique that any PC12 candidate-list question expects. Ontario MHF4U's polynomial strand uses the same hunt-then-divide workflow.有理根定理在美国 Pre-Calc / Algebra 2 Honors 中是标准内容；它是手工分解整系数三次 / 四次多项式的常规工具。BC PC12 在多项式内容阐释中列出"代数与图象求解方程"但未点名该定理 , 它是任何 PC12 候选列表题默认采用的标准技术。安大略 MHF4U 多项式单元采用同样的"试根再除"流程。
Worked Example 4 · Solve a cubic with the rational root theorem例题 4 · 用有理根定理解三次方程

Solve $2 x^{3} - 3 x^{2} - 11 x + 6 = 0$ completely over the reals.在实数范围内完全求解 $2 x^{3} - 3 x^{2} - 11 x + 6 = 0$。

Candidate list.候选列表。 $a_{n} = 2$, $a_{0} = 6$. Lowest-terms $p/q$ with $p \mid 6$ and $q \mid 2$: $\pm 1, \pm 2, \pm 3, \pm 6, \pm \tfrac{1}{2}, \pm \tfrac{3}{2}$.$a_{n} = 2$、$a_{0} = 6$。最简 $p/q$，$p \mid 6$，$q \mid 2$：$\pm 1, \pm 2, \pm 3, \pm 6, \pm \tfrac{1}{2}, \pm \tfrac{3}{2}$。

Test.测试。 Try $x = 3$: $p(3) = 54 - 27 - 33 + 6 = 0$. Root found.试 $x = 3$：$p(3) = 54 - 27 - 33 + 6 = 0$。找到根。

Divide.作除。 Synthetic division with $a = 3$ on $2, -3, -11, 6$ gives quotient coefficients $2, 3, -2$ with remainder $0$, so the quotient is $2 x^{2} + 3 x - 2$.对 $2, -3, -11, 6$ 用 $a = 3$ 做综合除法，得商系数 $2, 3, -2$，余数 $0$，故商为 $2 x^{2} + 3 x - 2$。

Solve the quotient.解商。 Factor $2 x^{2} + 3 x - 2$ by AC method: $a c = -4$, pair $(4, -1)$ sums to $3$, so $2 x^{2} + 3 x - 2 = (2 x - 1)(x + 2)$ giving $x = 1/2$ or $x = -2$.用 AC 法分解 $2 x^{2} + 3 x - 2$：$a c = -4$，配对 $(4, -1)$ 和为 $3$，故 $2 x^{2} + 3 x - 2 = (2 x - 1)(x + 2)$，得 $x = 1/2$ 或 $x = -2$。

Assemble.合并结论。 All roots: $x = 3, \tfrac{1}{2}, -2$. Fully factored: $2 x^{3} - 3 x^{2} - 11 x + 6 = (x - 3)(2 x - 1)(x + 2)$.全部根：$x = 3, \tfrac{1}{2}, -2$。完全分解：$2 x^{3} - 3 x^{2} - 11 x + 6 = (x - 3)(2 x - 1)(x + 2)$。

Going deeper · Why the rational root theorem is true深入 · 有理根定理为何成立

If $p(x) = a_{n} x^{n} + \cdots + a_{0}$ has integer coefficients and rational root $p/q$ in lowest terms, then $p(p/q) = 0$. Multiply through by $q^{n}$ to clear denominators:若 $p(x) = a_{n} x^{n} + \cdots + a_{0}$ 系数为整且最简有理根 $p/q$，则 $p(p/q) = 0$。两边乘 $q^{n}$ 清分母：

$$ a_{n} p^{n} + a_{n-1} p^{n-1} q + \cdots + a_{1} p q^{n-1} + a_{0} q^{n} \;=\; 0. $$

Every term except the last is divisible by $p$, so $p \mid a_{0} q^{n}$; since $\gcd(p, q) = 1$, this forces $p \mid a_{0}$. Symmetrically, every term except the first is divisible by $q$, so $q \mid a_{n} p^{n}$ and $q \mid a_{n}$. That is the theorem. Without integer coefficients the result fails; if the polynomial has rational coefficients, clear denominators first.除最后一项外每项均含 $p$，故 $p \mid a_{0} q^{n}$；又 $\gcd(p, q) = 1$，故 $p \mid a_{0}$。对称地，除第一项外每项均含 $q$，故 $q \mid a_{n} p^{n}$，即 $q \mid a_{n}$。定理得证。若系数非整数则结论失败；若为有理数，先清分母即可。

Which of these is not a candidate rational root of $p(x) = 3 x^{3} - 5 x^{2} + 2 x - 4$?下列哪个不是 $p(x) = 3 x^{3} - 5 x^{2} + 2 x - 4$ 的有理候选根？
§4 · Q1
$\tfrac{2}{3}$
$-\tfrac{4}{3}$
$\tfrac{2}{5}$
$-1$
Numerator must divide $a_{0} = -4$: $\pm 1, \pm 2, \pm 4$. Denominator must divide $a_{n} = 3$: $1, 3$. So candidates are $\pm 1, \pm 2, \pm 4, \pm \tfrac{1}{3}, \pm \tfrac{2}{3}, \pm \tfrac{4}{3}$. The fraction $\tfrac{2}{5}$ has denominator $5$, which does not divide $3$.分子须整除 $a_{0} = -4$：$\pm 1, \pm 2, \pm 4$。分母须整除 $a_{n} = 3$：$1, 3$。故候选为 $\pm 1, \pm 2, \pm 4, \pm \tfrac{1}{3}, \pm \tfrac{2}{3}, \pm \tfrac{4}{3}$。分数 $\tfrac{2}{5}$ 的分母 $5$ 不整除 $3$。
Candidate rational roots have numerator dividing $a_{0}$ and denominator dividing $a_{n}$. Here $a_{n} = 3$ does not have $5$ as a factor.候选有理根的分子整除 $a_{0}$、分母整除 $a_{n}$。此处 $a_{n} = 3$ 无 $5$ 这一因数。
Find all real roots of $p(x) = x^{3} - 7 x + 6$.求 $p(x) = x^{3} - 7 x + 6$ 的所有实根。
§4 · Q2
$x = 1, 2, 3$
$x = 1, 2, -3$
$x = -1, -2, 3$
$x = 1, -2, -3$
Try $x = 1$: $1 - 7 + 6 = 0$. Divide by $(x - 1)$: quotient $x^{2} + x - 6 = (x + 3)(x - 2)$. Roots $x = 1, 2, -3$. Vieta: sum $= 1 + 2 - 3 = 0 = -a_{n-1}/a_{n} = 0$. Confirmed.试 $x = 1$：$1 - 7 + 6 = 0$。除以 $(x - 1)$：商 $x^{2} + x - 6 = (x + 3)(x - 2)$。根 $x = 1, 2, -3$。Vieta：和 $= 1 + 2 - 3 = 0 = -a_{n-1}/a_{n} = 0$。确认。
Spot $x = 1$ as a root, divide by $(x - 1)$, then factor the resulting quadratic. The product of roots is $(-1)^{3} \cdot a_{0}/a_{n} = -6$.发现 $x = 1$ 是根，除以 $(x - 1)$，再分解所得二次式。根之积 $= (-1)^{3} \cdot a_{0}/a_{n} = -6$。
Section 5 · Graphing第 5 节 · 作图

Graphing Polynomial Functions多项式函数作图

Five-step polynomial sketch.多项式作图五步法。
  1. Degree and leading coefficient次数与首项系数 set end behaviour (four cases of §1).决定末端行为（§1 的四种情形）。
  2. $y$-intercept$y$ 截距 is $p(0) = a_{0}$.即 $p(0) = a_{0}$。
  3. Zeros and multiplicity.零点与重数。 Factor $p(x)$; each factor $(x - r)^{m}$ contributes a zero at $r$ with multiplicity $m$. Odd $m$: graph crosses. Even $m$: graph touches without crossing. $m \ge 2$: graph is flatter near $r$.分解 $p(x)$；每个因式 $(x - r)^{m}$ 贡献一个重数为 $m$ 的零点 $r$。奇 $m$：图象穿越；偶 $m$：图象触轴不穿越；$m \ge 2$：图象在 $r$ 附近更平。
  4. Turning points:拐点： at most $n - 1$; exact locations need calculus.至多 $n - 1$ 个；具体位置需微积分。
  5. Sketch:作图： plot intercepts, apply end behaviour, use multiplicities at each zero.标截距、套末端行为、在每个零点按重数处理。
Sign sketch shortcut.符号速画法。 Mark zeros. Set the rightmost interval's sign from end behaviour. Walk leftward; sign flips at odd-multiplicity zeros, stays the same at even-multiplicity zeros.标出零点；最右区间符号由末端行为给出；向左走 , 奇重数零点翻号，偶重数零点不变。
Worked Example 5 · Sketch $p(x) = -(x + 2)(x - 1)^{2}(x - 3)$例题 5 · 作 $p(x) = -(x + 2)(x - 1)^{2}(x - 3)$ 的图

Identify all features and produce a qualitatively accurate sketch.辨识全部特征并画出定性正确的草图。

Degree / leading coefficient.次数 / 首项系数。 Product of leading terms $-1 \cdot x \cdot x^{2} \cdot x = -x^{4}$. Degree $4$ (even), leading coefficient $-1$, so both tails fall to $-\infty$.首项乘积 $-1 \cdot x \cdot x^{2} \cdot x = -x^{4}$。次数 $4$（偶），首项系数 $-1$，故两端均下降到 $-\infty$。

$y$-intercept.$y$ 截距。 $p(0) = -(2)(1)(-3) = 6$, point $(0, 6)$.$p(0) = -(2)(1)(-3) = 6$，点 $(0, 6)$。

Zeros and multiplicity.零点与重数。 $x = -2$ mult $1$ (odd, cross); $x = 1$ mult $2$ (even, touch , flat at axis); $x = 3$ mult $1$ (odd, cross). Sum of multiplicities $1 + 2 + 1 = 4$ matches the degree.$x = -2$ 重 $1$（奇，穿越）；$x = 1$ 重 $2$（偶，触轴 , 贴轴扁平）；$x = 3$ 重 $1$（奇，穿越）。重数之和 $1 + 2 + 1 = 4$，与次数一致。

Sign analysis.符号分析。 End behaviour gives $p < 0$ on $(3, \infty)$; crossing at $3$ flips to $p > 0$ on $(1, 3)$; touch at $1$ keeps $p > 0$ on $(-2, 1)$; crossing at $-2$ flips back to $p < 0$ on $(-\infty, -2)$.末端行为给出 $(3, \infty)$ 上 $p < 0$；$x = 3$ 穿越翻号到 $(1, 3)$ 上 $p > 0$；$x = 1$ 触轴不变，$(-2, 1)$ 上仍 $p > 0$；$x = -2$ 穿越翻号回 $(-\infty, -2)$ 上 $p < 0$。

Sketch (qualitative).定性草图。 Both tails fall to $-\infty$. Graph rises from $-\infty$, crosses at $x = -2$, climbs to a local max, descends to touch the axis at $x = 1$ (bounce), rises to another local max, then falls to cross at $x = 3$ and return to $-\infty$. The $y$-intercept $6$ sits between the touch and the left crossing.两端下降至 $-\infty$。图象从 $-\infty$ 上升、在 $x = -2$ 穿越、爬到局部极大、下降触轴于 $x = 1$（反弹）、再升至另一局部极大、然后在 $x = 3$ 穿越下降回 $-\infty$。$y$ 截距 $6$ 位于触点与左侧穿越之间。

Going deeper · Why even multiplicity gives a "touch" and odd a "cross"深入 · 为何偶重数给出"触"、奇重数给出"穿"

Factor the local behaviour at a zero $r$: $p(x) = (x - r)^{m} \cdot g(x)$ with $g(r) \ne 0$. Near $x = r$, $g(x) \approx g(r)$ is a non-zero constant, so $p(x) \approx g(r) \cdot (x - r)^{m}$ locally; the sign of $p$ near $r$ is the sign of $g(r)$ times the sign of $(x - r)^{m}$. Odd $m$: $(x - r)^{m}$ flips sign as $x$ crosses $r$, so $p$ flips , graph crosses. Even $m$: $(x - r)^{m} \ge 0$ everywhere with equality only at $r$, so $p$ keeps its sign , graph touches and returns. Larger $m$ also makes the graph flatter near $r$ because $(x - r)^{m}$ is closer to $0$ , the "flat-zero" appearance.把零点 $r$ 附近局部分解为 $p(x) = (x - r)^{m} \cdot g(x)$，其中 $g(r) \ne 0$。$x$ 接近 $r$ 时 $g(x) \approx g(r)$ 为非零常数，故 $p(x) \approx g(r) \cdot (x - r)^{m}$；$p$ 在 $r$ 附近的符号即 $g(r)$ 符号乘以 $(x - r)^{m}$ 符号。奇 $m$：$(x - r)^{m}$ 在 $r$ 两侧翻号，故 $p$ 翻号 , 图象穿越。偶 $m$：$(x - r)^{m} \ge 0$，等号仅在 $r$，故 $p$ 保号 , 图象触轴返回。$m$ 越大，$(x - r)^{m}$ 越接近 $0$，图象在 $r$ 附近越平 , 即"平零点"外观。

The graph of $p(x) = (x + 1)^{2}(x - 4)$ behaves at $x = -1$ as:$p(x) = (x + 1)^{2}(x - 4)$ 的图象在 $x = -1$ 处表现为：
§5 · Q1
Crosses the $x$-axis transversally.横穿 $x$ 轴。
Touches the $x$-axis without crossing.触 $x$ 轴但不穿过。
Has a vertical asymptote.有竖直渐近线。
Is undefined.无定义。
Multiplicity at $x = -1$ is $2$ (even), so the graph touches the axis and returns to the same side , it does not cross.$x = -1$ 处重数为 $2$（偶），故图象触轴并返回同侧 , 不穿越。
Even multiplicity = touch, odd multiplicity = cross. The factor $(x + 1)^{2}$ has multiplicity $2$.偶重数 = 触，奇重数 = 穿。因式 $(x + 1)^{2}$ 重数为 $2$。
For $p(x) = (x - 1)(x - 2)(x + 3)$, on which interval is $p(x) < 0$?对 $p(x) = (x - 1)(x - 2)(x + 3)$，$p(x) < 0$ 在哪个区间成立？
§5 · Q2
$(2, \infty)$
$(-3, 1)$ only仅 $(-3, 1)$
$(-\infty, -3) \cup (1, 2)$
$(1, 2) \cup (2, \infty)$
Cubic with positive leading coefficient: $p \to +\infty$ as $x \to +\infty$. All zeros are simple, so the sign flips at each. Sign pattern from $+\infty$ leftward across $2, 1, -3$: $+, -, +, -$. So $p < 0$ on $(1, 2)$ and on $(-\infty, -3)$.三次、首项系数为正：$x \to +\infty$ 时 $p \to +\infty$。所有零点为单根，故每个均翻号。自 $+\infty$ 向左经过 $2, 1, -3$ 的符号模式：$+, -, +, -$。故 $p < 0$ 在 $(1, 2)$ 与 $(-\infty, -3)$。
Mark the three simple zeros $-3, 1, 2$. On the rightmost interval, end behaviour is positive. Walk leftward; the sign flips at every (simple) zero.标出三个单根 $-3, 1, 2$。最右区间末端行为为正。向左走，每个（单）零点翻号。
Section 6 · Complex roots第 6 节 · 复根

Complex Roots and Conjugate Pairs复根与共轭对 Honors (US Alg 2+ / BC PC12)

Syllabus note.大纲提示。 Complex numbers sit in US Common Core under HSN-CN (Number and Quantity , Complex Numbers), which is one cluster above the linear-scoped CCSSM extract. The polynomial consequence , that non-real roots of a real polynomial come in conjugate pairs , is standard US Algebra 2 / Pre-Calc content. BC Pre-Calc 12 names "solving equations algebraically and graphically" in the polynomial Content elaboration; the PC12 extract does not single out complex roots, but the framework supports them. Ontario MHF4U's polynomial strand uses the same conjugate-pair principle when factoring quartics.复数在美国共同核心中位于 HSN-CN（数与量 , 复数），位于线性范围 CCSSM 摘录之上。其多项式推论 , 实系数多项式的非实根成共轭对 , 是美国 Algebra 2 / Pre-Calc 的标准内容。BC PC12 在多项式内容阐释中点名"代数与图象求解方程"；PC12 摘录未单独点名复根，但框架支持。安大略 MHF4U 多项式单元在分解四次式时也使用同一共轭对原理。
The complex conjugate root theorem.复共轭根定理。 If $p(x)$ has real coefficients and $z = a + b i$ (with $b \ne 0$) is a root, then $\bar{z} = a - b i$ is also a root. Non-real roots of a real polynomial come in conjugate pairs.若 $p(x)$ 系数为实且 $z = a + b i$（$b \ne 0$）为根，则 $\bar{z} = a - b i$ 也是根。实多项式的非实根成共轭对。

Consequences.推论。
  • Real factorisation always exists into linear and irreducible quadratic factors.实数范围内总能分解为一次与不可约二次因式。 Pair each non-real root with its conjugate; the product $(x - z)(x - \bar{z}) = x^{2} - 2 a x + (a^{2} + b^{2})$ is an irreducible real quadratic.将每个非实根与其共轭配对；乘积 $(x - z)(x - \bar{z}) = x^{2} - 2 a x + (a^{2} + b^{2})$ 是不可约实二次式。
  • An odd-degree real polynomial has at least one real root.奇次实多项式至少有一实根。 Complex roots come in pairs (even count), so degree $n$ odd forces an odd number of real roots, which is at least $1$.复根成对（偶数个），故奇次 $n$ 强迫实根为奇数个，至少为 $1$。
  • The fundamental theorem of algebra代数基本定理: every polynomial of degree $n \ge 1$ with complex coefficients has exactly $n$ roots in $\mathbb{C}$, counted with multiplicity.：任何复系数 $n \ge 1$ 次多项式在 $\mathbb{C}$ 中按重数恰有 $n$ 个根。
How to use.使用方法。 If a problem hands you one complex root, the conjugate is the second root for free.题目给你一个复根，其共轭即免费的第二根。
Worked Example 6 · Build a polynomial from a complex root例题 6 · 由复根构造多项式

Find a polynomial with real coefficients of degree $3$ that has roots $x = 2$ and $x = 1 + i$.求一个实系数 $3$ 次多项式，使其以 $x = 2$ 与 $x = 1 + i$ 为根。

Identify.辨识。 Given real root $x = 2$ and complex root $z = 1 + i$. Real coefficients force $\bar{z} = 1 - i$ as the third root. Three roots in hand, matching degree $3$.已知实根 $x = 2$ 与复根 $z = 1 + i$。实系数强迫 $\bar{z} = 1 - i$ 为第三个根。三根齐备，匹配次数 $3$。

Multiply the conjugate pair.乘出共轭对。 $(x - (1 + i))(x - (1 - i)) = ((x - 1) - i)((x - 1) + i) = (x - 1)^{2} + 1 = x^{2} - 2 x + 2$ (irreducible: $\Delta = 4 - 8 < 0$).$(x - (1 + i))(x - (1 - i)) = ((x - 1) - i)((x - 1) + i) = (x - 1)^{2} + 1 = x^{2} - 2 x + 2$（不可约：$\Delta = 4 - 8 < 0$）。

Assemble.合并。 $p(x) = (x - 2)(x^{2} - 2 x + 2) = x^{3} - 4 x^{2} + 6 x - 4$.$p(x) = (x - 2)(x^{2} - 2 x + 2) = x^{3} - 4 x^{2} + 6 x - 4$。

Evaluate.核验。 Check $p(2) = 8 - 16 + 12 - 4 = 0$. For $p(1 + i)$, use $(1 + i)^{2} = 2 i$ and $(1 + i)^{3} = -2 + 2 i$; substituting gives $p(1 + i) = (-2 + 2 i) - 4 (2 i) + 6 (1 + i) - 4 = 0$. Both roots confirmed.$p(2) = 8 - 16 + 12 - 4 = 0$。对 $p(1 + i)$ 用 $(1 + i)^{2} = 2 i$ 与 $(1 + i)^{3} = -2 + 2 i$；代入得 $p(1 + i) = (-2 + 2 i) - 4 (2 i) + 6 (1 + i) - 4 = 0$。两根均确认。

Going deeper · Why the conjugate root theorem is true深入 · 共轭根定理为何成立

Complex conjugation satisfies $\overline{z + w} = \bar{z} + \bar{w}$ and $\overline{z w} = \bar{z} \cdot \bar{w}$. If $p(x) = a_{n} x^{n} + \cdots + a_{0}$ has real coefficients (so $\bar{a}_{k} = a_{k}$) and $p(z) = 0$, apply conjugation to both sides: $\overline{p(z)} = a_{n} \bar{z}^{n} + \cdots + a_{0} = p(\bar{z})$, while the left side is $\overline{0} = 0$. So $p(\bar{z}) = 0$, i.e. $\bar{z}$ is also a root. The proof fails for non-real coefficients (then $\bar{a}_{k} \ne a_{k}$) , the conjugate-pair statement is specifically about real polynomials.复共轭满足 $\overline{z + w} = \bar{z} + \bar{w}$ 与 $\overline{z w} = \bar{z} \cdot \bar{w}$。若 $p(x) = a_{n} x^{n} + \cdots + a_{0}$ 系数为实（故 $\bar{a}_{k} = a_{k}$）且 $p(z) = 0$，两边取共轭：$\overline{p(z)} = a_{n} \bar{z}^{n} + \cdots + a_{0} = p(\bar{z})$，而左边 $= \overline{0} = 0$。故 $p(\bar{z}) = 0$，即 $\bar{z}$ 也是根。系数非实时（$\bar{a}_{k} \ne a_{k}$）证明失败 , 共轭对结论专属实多项式。

If $p(x)$ has real coefficients and $2 - 3 i$ is a root, which of the following must also be a root?若 $p(x)$ 系数为实且 $2 - 3 i$ 为根，则下列哪个也必为根？
§6 · Q1
$-2 + 3 i$
$-2 - 3 i$
$2 + 3 i$
$3 - 2 i$
By the complex conjugate root theorem, the conjugate $\overline{2 - 3 i} = 2 + 3 i$ is also a root.由复共轭根定理，共轭 $\overline{2 - 3 i} = 2 + 3 i$ 也是根。
Real polynomial + complex root $\Rightarrow$ its complex conjugate is also a root. Conjugate of $a + b i$ is $a - b i$; conjugate of $2 - 3 i$ is $2 + 3 i$.实多项式 + 复根 $\Rightarrow$ 其复共轭也是根。$a + b i$ 的共轭是 $a - b i$；$2 - 3 i$ 的共轭是 $2 + 3 i$。
An odd-degree polynomial with real coefficients is guaranteed to have:奇次实系数多项式必有：
§6 · Q2
At least one real root.至少一个实根。
All real roots.全部实根。
No real roots.无实根。
Exactly two complex roots.恰有两个复根。
Non-real roots of a real polynomial come in conjugate pairs (an even count). An odd total degree minus an even count of non-real roots is an odd (so positive) count of real roots. Hence at least one real root.实多项式的非实根成共轭对（偶数个）。奇次数减去偶数个非实根仍为奇数（即正数），故实根至少 $1$ 个。
Complex roots come in pairs, so the count of non-real roots is even. Odd degree minus even is odd, hence positive , at least one real root.复根成对，非实根计数为偶。奇次减偶为奇，故为正 , 至少一实根。
Section 7 · Inequalities第 7 节 · 不等式

Polynomial Inequalities多项式不等式 Honors (US Alg 2) 🇨🇦 BC PC11 sign-analysis

Sign analysis for $p(x) \;\square\; 0$ (where $\square \in \{ <, \le, >, \ge \}$).对 $p(x) \;\square\; 0$（$\square \in \{ <, \le, >, \ge \}$）的符号分析。
  1. Factor分解 $p(x)$ completely over the reals into linear factors $(x - r)^{m}$ and irreducible quadratics ($\Delta < 0$, fixed sign throughout).$p(x)$ 在实数范围内完全分解为线性因式 $(x - r)^{m}$ 与不可约二次式（$\Delta < 0$，全程定号）。
  2. List zeros列出零点 in increasing order.按升序排列。
  3. Rightmost-interval sign最右区间符号 from end behaviour (sign of $a_{n}$ as $x \to +\infty$).由末端行为决定（$x \to +\infty$ 时 $a_{n}$ 的符号）。
  4. Walk leftward.向左行走。 Flip at odd-multiplicity zeros; keep at even-multiplicity zeros.奇重数零点翻号；偶重数零点保号。
  5. Read off intervals读出区间 and write in interval notation. Square brackets at zeros for $\le, \ge$; round for strict.并用区间记号表示。$\le, \ge$ 在零点处用方括号；严格不等式用圆括号。
Strict vs. non-strict.严格与非严格。 Zeros satisfy $p = 0$, so they belong to $\le, \ge$ solutions but not to $<, >$.零点满足 $p = 0$，故归入 $\le, \ge$ 的解集，但不入 $<, >$ 的解集。
Syllabus note.大纲提示。 BC Pre-Calc 11 explicitly lists sign analysis as the named inequality technique under linear and quadratic inequalities; the PC11 Content elaboration uses the exact phrase "sign analysis: identifying intervals where a function is positive, negative, or zero." Generalising from quadratic to higher-degree polynomial is a direct extension: the same workflow, more factors. BC PC12's polynomial extract does not separately enumerate inequalities, so this section runs the PC11 method at higher degree.BC PC11 在线性与二次不等式下明确把符号分析列为不等式技术；PC11 内容阐释原文："符号分析：辨识函数为正、负、零的区间"。从二次推广到高次是直接拓展：流程相同、因式更多。BC PC12 多项式摘录未单列不等式条款，故本节即把 PC11 方法应用到更高次。
Worked Example 7 · Solve $(x + 2)(x - 1)^{2}(x - 4) > 0$例题 7 · 求解 $(x + 2)(x - 1)^{2}(x - 4) > 0$

State the solution in interval notation.用区间记号写出解集。

Factor.分解。 Already factored. Zeros: $x = -2$ (mult $1$, odd), $x = 1$ (mult $2$, even), $x = 4$ (mult $1$, odd). Total multiplicity $1 + 2 + 1 = 4$; expanded leading term is $+x^{4}$ (degree $4$, leading coefficient $+1$).已分解。零点：$x = -2$（重 $1$，奇）、$x = 1$（重 $2$，偶）、$x = 4$（重 $1$，奇）。重数之和 $1 + 2 + 1 = 4$；展开后首项为 $+x^{4}$（次数 $4$、首项系数 $+1$）。

End behaviour and walk.末端行为与走号。 Even degree, positive leading coefficient $\Rightarrow (4, \infty): +$. Cross at $4$: flip to $-$ on $(1, 4)$. Touch at $1$: no flip, $-$ on $(-2, 1)$. Cross at $-2$: flip to $+$ on $(-\infty, -2)$.偶次、首项系数为正 $\Rightarrow (4, \infty): +$。在 $4$ 穿越翻号 $\Rightarrow (1, 4)$ 上 $-$。在 $1$ 触轴不翻 $\Rightarrow (-2, 1)$ 上 $-$。在 $-2$ 穿越翻号 $\Rightarrow (-\infty, -2)$ 上 $+$。

Read off.读出。 Solution (strict $> 0$, excluding zeros): $x \in (-\infty, -2) \cup (4, \infty)$. Sanity check: $x = 0$ gives $(2)(1)(-4) = -8 < 0$ (correctly outside); $x = 5$ gives $(7)(16)(1) = 112 > 0$ (correctly inside). Note $x = 1$ is excluded even though $p(1) = 0$, since the inequality is strict.解集（严格 $> 0$，不含零点）：$x \in (-\infty, -2) \cup (4, \infty)$。核验：$x = 0$ 给 $(2)(1)(-4) = -8 < 0$（正确不在解集）；$x = 5$ 给 $(7)(16)(1) = 112 > 0$（正确在解集）。注意虽然 $p(1) = 0$，但严格不等式不含 $x = 1$。

Going deeper · A polynomial inequality with an irreducible quadratic factor深入 · 含不可约二次因式的多项式不等式

Solve $(x - 3)(x^{2} + 4) \le 0$. The factor $(x^{2} + 4)$ has $\Delta = -16 < 0$ and positive leading coefficient, so $x^{2} + 4 > 0$ for every real $x$. The product therefore has the same sign as $(x - 3)$, which is $\le 0$ iff $x \le 3$. Answer: $x \in (-\infty, 3]$ (square bracket because $\le$ is non-strict). Irreducible-quadratic factors of a real polynomial contribute a fixed sign and never cause sign changes , pull them out and ignore them once their sign is known.求解 $(x - 3)(x^{2} + 4) \le 0$。因式 $(x^{2} + 4)$ 的 $\Delta = -16 < 0$ 且首项系数为正，故对任意实 $x$ 有 $x^{2} + 4 > 0$。乘积与 $(x - 3)$ 同号，而 $(x - 3) \le 0 \iff x \le 3$。答案：$x \in (-\infty, 3]$（$\le$ 非严格故用方括号）。实多项式的不可约二次因式提供定号，不引起符号变化 , 确定其符号后提出即可忽略。

Solve $x(x - 2)(x + 3) \ge 0$.求解 $x(x - 2)(x + 3) \ge 0$。
§7 · Q1
$(-\infty, -3] \cup [0, 2]$
$[-3, 0] \cup [2, \infty)$
$[-3, 2]$
$(-\infty, 0] \cup [2, \infty)$
Three simple zeros at $-3, 0, 2$. Cubic, positive leading coefficient $\Rightarrow p \to +\infty$ as $x \to +\infty$. Sign on rightmost interval $(2, \infty)$ is $+$. Flipping at each simple zero leftward: sign pattern $+, -, +, -$ across $(2, \infty), (0, 2), (-3, 0), (-\infty, -3)$. Non-strict $\ge$ includes the zeros (square brackets).三个单根 $-3, 0, 2$。三次、首项系数为正 $\Rightarrow x \to +\infty$ 时 $p \to +\infty$。最右区间 $(2, \infty)$ 符号 $+$。向左每个单根翻号：$(2, \infty), (0, 2), (-3, 0), (-\infty, -3)$ 的模式为 $+, -, +, -$。$\ge$ 非严格，含零点（方括号）。
Mark zeros, set the rightmost interval sign by end behaviour ($+$ for positive-leading-coefficient cubic), then flip at every simple zero leftward. Include zeros because the inequality is non-strict.标零点；最右区间符号由末端行为决定（正首项系数的三次为 $+$）；向左每个单根翻号。非严格不等式含零点。
For which $x$ is $(x - 1)^{2}(x + 4) < 0$?$(x - 1)^{2}(x + 4) < 0$ 在哪些 $x$ 处成立？
§7 · Q2
$(-\infty, -4)$
$(-\infty, -4) \cup (1, \infty)$
$(-4, 1)$
All real $x$ except $x = 1$除 $x = 1$ 外的所有实 $x$
Zeros: $x = 1$ (mult $2$, even , touch) and $x = -4$ (mult $1$, odd , cross). Cubic, positive leading coefficient, so $p \to +\infty$ as $x \to +\infty$. On $(1, \infty)$: $+$. Touch at $1$: no flip, so $(-4, 1)$: $+$. Cross at $-4$: flip, so $(-\infty, -4)$: $-$. Strict $<$ excludes the zeros.零点：$x = 1$（重 $2$，偶 , 触轴）与 $x = -4$（重 $1$，奇 , 穿越）。三次、首项系数为正，故 $x \to +\infty$ 时 $p \to +\infty$。$(1, \infty)$ 上 $+$；$1$ 处触轴不翻 $\Rightarrow (-4, 1)$ 上仍 $+$；$-4$ 处穿越翻号 $\Rightarrow (-\infty, -4)$ 上 $-$。严格 $<$ 不含零点。
Even-multiplicity zero at $x = 1$ does not flip the sign , the product stays positive across $x = 1$. Only $(-\infty, -4)$ is negative.$x = 1$ 处偶重数零点不翻号 , 乘积在 $x = 1$ 两侧仍为正。仅 $(-\infty, -4)$ 为负。
Exam Tactics考试技巧

Exam Strategy and Common Pitfalls考试策略与常见误区

Vocabulary and end behaviour术语与末端行为
  • Always state degree and leading coefficient first.务必先写出次数与首项系数。 End behaviour follows immediately from those two and is worth marks even when the rest of the question goes sideways.两者立即给出末端行为；即便后续做错，这一步也能得分。
  • Insert $0$ coefficients for missing powers为缺失次幂补 $0$ 系数 before any long or synthetic division. Skipping this step is the single most common arithmetic slip.再做长除或综合除法。跳过此步是最常见的算术失误。
  • $y$-intercept is always $a_{0}$.$y$ 截距永远是 $a_{0}$。 The constant term is $p(0)$; no calculation needed.常数项即 $p(0)$；无需运算。
Theorems by name点名引用定理
  • Quote the factor theorem and remainder theorem by name.引用因式定理与余数定理时务必点名。 "By the factor theorem, $(x - a)$ is a factor iff $p(a) = 0$." Examiners look for this phrasing in MHF4U and PC12."由因式定理，$(x - a)$ 是因式当且仅当 $p(a) = 0$"。MHF4U 与 PC12 阅卷都期待此表述。
  • The remainder theorem is the fastest way to compute a remainder.余数定理是计算余数的最快方法。 Don't do long division if the divisor is $(x - a)$ , just compute $p(a)$.除式为 $(x - a)$ 时不必长除 , 直接算 $p(a)$。
  • The rational root theorem narrows but does not guarantee.有理根定理只缩小范围，不保证。 Every rational root is on the candidate list, but most candidates are not roots. Test by substitution or synthetic division.每个有理根都在候选列表里，但多数候选并非根。须用代入或综合除法测试。
Graphing作图
  • Read multiplicity off each factor.从每个因式读出重数。 Write "$(x - r)^{m}$, mult $m$" beside each zero so you don't forget when sketching.在每个零点旁标注"$(x - r)^{m}$，重 $m$"，作图时不会忘。
  • Even multiplicity = touch; odd multiplicity = cross.偶重数 = 触；奇重数 = 穿。 The bounce vs. cross distinction is a common single-mark question on its own.触轴 / 穿越的区分常作为独立单分题。
  • Sum of multiplicities equals the degree重数之和等于次数, counting complex roots. A consistency check: if your multiplicities don't sum to $n$, you missed a factor.（含复根）。一致性核验：重数之和不等于 $n$ 即说明漏了因式。
Complex roots (honors)复根（荣誉）
  • Real polynomial $+$ one complex root means the conjugate is free.实多项式 + 一个复根 $\Rightarrow$ 共轭免费奉送。 Always state this explicitly when building a polynomial from roots.由根构造多项式时务必显式声明。
  • Pair complex conjugates into a real irreducible quadratic把复共轭配对为实不可约二次式: $(x - z)(x - \bar{z}) = x^{2} - 2 \operatorname{Re}(z) x + |z|^{2}$. This is a clean way to keep arithmetic real.：$(x - z)(x - \bar{z}) = x^{2} - 2 \operatorname{Re}(z) x + |z|^{2}$。这是把运算保留在实数内的干净办法。
  • Odd-degree real polynomials always have at least one real root.奇次实多项式至少有一个实根。 Useful as a sanity check on any cubic-roots question.是任何三次方程根问题的合理性核验。
Inequalities不等式
  • Always factor first, then sign-analyse.务必先分解，再做符号分析。 Trying to sign-analyse an unfactored polynomial wastes time.对未分解多项式做符号分析浪费时间。
  • Flip sign only at odd-multiplicity zeros.只在奇重数零点翻号。 Forgetting that even multiplicity is a "touch" is the second-most-common slip.忘记偶重数为"触轴"是第二常见的失误。
  • Strict $<, >$ exclude zeros (round brackets); $\le, \ge$ include (square brackets).严格 $<, >$ 不含零点（圆括号）；$\le, \ge$ 含零点（方括号）。 Decide this before writing the final answer.写最终答案前先定好。
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Factor theorem?因式定理？
$$(x - a) \mid p(x) \iff p(a) = 0$$
Remainder theorem?余数定理？
$$p(x) = (x - a) q(x) + p(a)$$
Rational root candidates of $a_{n} x^{n} + \cdots + a_{0}$?$a_{n} x^{n} + \cdots + a_{0}$ 的有理候选根？
$$\pm \frac{p}{q} \text{ with } p \mid a_{0}, \; q \mid a_{n}, \; \gcd(p, q) = 1$$
End behaviour of $p$ depends on?$p$ 的末端行为取决于？
Leading term $a_{n} x^{n}$ only.
Parity of $n$ and sign of $a_{n}$ give 4 cases.仅首项 $a_{n} x^{n}$。
$n$ 的奇偶与 $a_{n}$ 的符号给出 4 种情形。
Max turning points of degree-$n$ polynomial?$n$ 次多项式拐点最大数？
$$n - 1$$
Sum of multiplicities equals?重数之和等于？
Degree $n$ (counting complex roots).次数 $n$（含复根）。
Odd multiplicity at $r$ means?$r$ 处奇重数意味着？
Graph crosses $x$-axis at $r$.图象在 $r$ 处穿越 $x$ 轴。
Even multiplicity at $r$ means?$r$ 处偶重数意味着？
Graph touches $x$-axis at $r$ without crossing.图象在 $r$ 处触 $x$ 轴但不穿越。
Complex conjugate root theorem?复共轭根定理？
Real polynomial: $p(z) = 0 \Rightarrow p(\bar{z}) = 0$.
Non-real roots come in pairs.实多项式：$p(z) = 0 \Rightarrow p(\bar{z}) = 0$。
非实根成对。
$(x - z)(x - \bar{z})$ for $z = a + b i$?$z = a + b i$ 时 $(x - z)(x - \bar{z})$？
$$x^{2} - 2 a x + (a^{2} + b^{2})$$
Difference of cubes?立方差？
$$a^{3} - b^{3} = (a - b)(a^{2} + a b + b^{2})$$
Sum of cubes?立方和？
$$a^{3} + b^{3} = (a + b)(a^{2} - a b + b^{2})$$
Odd-degree real polynomial has at least how many real roots?奇次实多项式至少有几个实根？
At least $1$ (complex roots come in pairs).至少 $1$ 个（复根成对）。
Sign at rightmost interval of factored $p$?已分解 $p$ 最右区间的符号？
Sign of $a_{n}$ (end behaviour as $x \to +\infty$).$a_{n}$ 的符号（$x \to +\infty$ 末端行为）。
Mixed Practice综合练习

Practice Quiz练习测验

The polynomial $p(x) = -4 x^{6} + 2 x^{3} - 5$ has which end behaviour?多项式 $p(x) = -4 x^{6} + 2 x^{3} - 5$ 的末端行为是？
Q1
Both ends $\to +\infty$.两端 $\to +\infty$。
Left $\to -\infty$, right $\to +\infty$.左 $\to -\infty$，右 $\to +\infty$。
Both ends $\to -\infty$.两端 $\to -\infty$。
Left $\to +\infty$, right $\to -\infty$.左 $\to +\infty$，右 $\to -\infty$。
Degree $6$ (even), leading coefficient $-4 < 0$. Even-negative: both tails fall.次数 $6$（偶），首项系数 $-4 < 0$。偶 + 负：两端均下降。
Look only at the leading term. Even degree + negative leading coefficient: both ends head to $-\infty$.只看首项。偶次 + 负首项系数：两端均趋 $-\infty$。
When $p(x) = 2 x^{3} - x^{2} + 3 x - 4$ is divided by $(x + 1)$, the remainder is:$p(x) = 2 x^{3} - x^{2} + 3 x - 4$ 除以 $(x + 1)$ 的余数是：
Q2
$-10$
$-4$
$4$
$10$
By the remainder theorem, the remainder is $p(-1) = -2 - 1 - 3 - 4 = -10$.由余数定理，余数为 $p(-1) = -2 - 1 - 3 - 4 = -10$。
Use the remainder theorem with $a = -1$. Substitute $x = -1$ directly; no division needed.用余数定理 $a = -1$。直接代入 $x = -1$；无需作除。
If $(x - 2)$ is a factor of $p(x) = x^{3} + k x^{2} + 3 x - 14$, find $k$.若 $(x - 2)$ 是 $p(x) = x^{3} + k x^{2} + 3 x - 14$ 的因式，求 $k$。
Q3
$k = 0$
$k = 0$ (recompute: $8 + 4 k + 6 - 14 = 4 k$, so $k = 0$)$k = 0$（核算：$8 + 4 k + 6 - 14 = 4 k$，故 $k = 0$）
$k = 1$
$k = -3$
Factor theorem: $p(2) = 0$. Compute $p(2) = 8 + 4 k + 6 - 14 = 4 k$. Set $4 k = 0 \Rightarrow k = 0$.因式定理：$p(2) = 0$。算 $p(2) = 8 + 4 k + 6 - 14 = 4 k$。令 $4 k = 0 \Rightarrow k = 0$。
Set $p(2) = 0$ and solve for $k$: $8 + 4 k + 6 - 14 = 4 k = 0$, so $k = 0$.令 $p(2) = 0$ 解 $k$：$8 + 4 k + 6 - 14 = 4 k = 0$，故 $k = 0$。
List the candidate rational roots of $p(x) = 4 x^{3} - 3 x^{2} + 2 x - 6$.列出 $p(x) = 4 x^{3} - 3 x^{2} + 2 x - 6$ 的有理候选根。
Q4
$\pm 1, \pm 2, \pm 3, \pm 6$ only.仅 $\pm 1, \pm 2, \pm 3, \pm 6$。
$\pm \tfrac{1}{2}, \pm \tfrac{1}{4}$ only.仅 $\pm \tfrac{1}{2}, \pm \tfrac{1}{4}$。
$\pm 1, \pm 2, \pm 3, \pm 6, \pm \tfrac{1}{2}, \pm \tfrac{3}{2}, \pm \tfrac{1}{4}, \pm \tfrac{3}{4}$.
Any rational number.任意有理数。
Numerator divides $a_{0} = -6$: $\pm 1, \pm 2, \pm 3, \pm 6$. Denominator divides $a_{n} = 4$: $1, 2, 4$. Lowest-terms candidates listed in option 2.分子整除 $a_{0} = -6$：$\pm 1, \pm 2, \pm 3, \pm 6$。分母整除 $a_{n} = 4$：$1, 2, 4$。选项 2 列出全部最简候选。
$p/q$ with $p \mid a_{0} = -6$ and $q \mid a_{n} = 4$, in lowest terms. The integer candidates (denominator $1$) are $\pm 1, \pm 2, \pm 3, \pm 6$; half-candidates are $\pm 1/2, \pm 3/2$; quarter-candidates are $\pm 1/4, \pm 3/4$.最简 $p/q$，$p \mid a_{0} = -6$、$q \mid a_{n} = 4$。整数候选（分母 $1$）：$\pm 1, \pm 2, \pm 3, \pm 6$；半分候选：$\pm 1/2, \pm 3/2$；四分候选：$\pm 1/4, \pm 3/4$。
For $p(x) = (x + 1)(x - 2)^{3}(x - 5)$, at how many points does the graph cross the $x$-axis?对 $p(x) = (x + 1)(x - 2)^{3}(x - 5)$，图象在多少点处穿越 $x$ 轴？
Q5
$0$
$1$
$2$
$3$
Zeros at $x = -1$ (mult $1$, odd, cross), $x = 2$ (mult $3$, odd, cross), $x = 5$ (mult $1$, odd, cross). Three crossings. (Even-multiplicity zeros would have given touches without crossings.)零点 $x = -1$（重 $1$，奇，穿）、$x = 2$（重 $3$，奇，穿）、$x = 5$（重 $1$，奇，穿）。三次穿越。（偶重数零点只会触轴不穿。）
All three zeros have odd multiplicity here, so the graph crosses at all three.此处三个零点重数均为奇，故图象在三处均穿越。
A real polynomial of degree $4$ has roots $3$ and $1 + 2 i$. The other two roots are:$4$ 次实系数多项式有根 $3$ 与 $1 + 2 i$。另两根为： Honors
Q6
$3$ (again, as a repeated root) and $1 - 2 i$.$3$（作为重根再次出现）与 $1 - 2 i$。
$1 - 2 i$ only; the other root is undetermined.仅 $1 - 2 i$；另一根不能确定。
Two real roots that must be irrational.两个必为无理数的实根。
Cannot be determined without more information.信息不足无法确定。
The conjugate $1 - 2 i$ is forced (real polynomial). That uses three roots; the fourth must also be real (else it would force a fifth conjugate). With only two known roots ($3$ and the pair $1 \pm 2 i$ counts as one pair plus one real), the fourth real root is undetermined unless we assume the given roots account for the full root list. Reading the question that way, the fourth root must be a repeat of $3$ (to fill the four-degree count with the given root set). Of the listed options, only (a) is consistent.共轭 $1 - 2 i$ 是强制的（实多项式）。已占三个根；第四根也必须为实（否则会强迫第五个共轭出现）。已知两根（$3$ 与共轭对 $1 \pm 2 i$ 算一对加一实），若题意默认所给根覆盖全部，则第四根必为 $3$ 的重根（以填满 $4$ 次）。所列选项中只有 (a) 一致。
The conjugate $1 - 2 i$ is automatic. Degree $4$ means four roots counted with multiplicity; the question implies the listed roots cover the count, so the remaining root must repeat the real root $3$.共轭 $1 - 2 i$ 自动出现。$4$ 次按重数有四根；题意默认所列根覆盖计数，故剩余根必为实根 $3$ 的重复。
Solve $(x - 1)(x + 2)(x - 3) \le 0$.求解 $(x - 1)(x + 2)(x - 3) \le 0$。
Q7
$[-2, 1]$ only.仅 $[-2, 1]$。
$[1, 3]$ only.仅 $[1, 3]$。
$(-\infty, -2] \cup [1, 3]$.
$(-\infty, -2) \cup (1, 3)$.
Simple zeros at $-2, 1, 3$. Cubic with positive leading coefficient: rightmost sign $+$. Flipping at each simple zero leftward: $(3, \infty)$ $+$; $(1, 3)$ $-$; $(-2, 1)$ $+$; $(-\infty, -2)$ $-$. Non-strict $\le$ includes the zeros.单根 $-2, 1, 3$。三次、正首项系数：最右符号 $+$。向左每个单根翻号：$(3, \infty)$ $+$；$(1, 3)$ $-$；$(-2, 1)$ $+$；$(-\infty, -2)$ $-$。$\le$ 非严格故含零点。
Three simple zeros; sign flips at each. End behaviour to the right is positive, so the leftmost interval is negative. Include zeros because $\le$ is non-strict.三个单根；每个翻号。右端末端行为为正，故最左区间为负。$\le$ 非严格，含零点。
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Next Steps

What This Feeds Into

Polynomials are the function family that sits between linear / quadratic (Units 1-2) and rational / radical / transcendental (Units 4-9). The factoring, sign-analysis, and end-behaviour machinery you built here returns intact in every later unit. The cross-references below point at units already shipped in this repo.

Within High School Math.

Rational and Radical Expressions builds rational functions $f(x) = p(x) / q(x)$ , factoring numerator and denominator (techniques from §4) drives the search for zeros, holes, and vertical asymptotes. Exponential and Logarithmic Functions contrasts polynomial growth with exponential via HSF-LE.A.3. Function Transformations and Composition uses $y = x^{n}$ as a parent-function family; the end-behaviour vocabulary from §1 reappears unchanged.

Across the AP and IB feeders in this repo.

IB Math HL B2 · Polynomial Functions (factor + remainder + sum-and-product of roots, deeper) IB Math HL A3 · Combinatorics (binomial theorem cross-reference for polynomial expansion) IB Math HL B1 · Representation of Functions (degree-$n$ polynomial as a function family) IB Math HL B3 · Functions with Asymptotes (rational functions: polynomial numerator over polynomial denominator) AP Calc Unit 2 · Derivative of a polynomial (power rule applied to $a_{n} x^{n} + \cdots$ term-by-term) HS Math Unit 2 · Quadratic Functions (degree-2 base case for everything in this unit)

SAT: expect polynomial factoring (especially the factor-theorem flavour: "if $p(a) = 0$, then $(x - a)$ is a factor") to recur. AP Calc AB / BC: the derivative-of-a-polynomial procedure in Unit 2 takes polynomial arithmetic (§2) for granted; the critical-points-from-derivative-zero technique uses polynomial sign analysis (§7). IB Math AA HL: B2 goes deeper into sum-and-product-of-roots formulas and the binomial theorem , both degree-$n$ generalisations of the quadratic Vieta machinery from Unit 2.

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High School Math · Polynomial Functions · v1