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Polynomial Functions · Solutions多项式函数 · 详解

Companion to the Practice Set · Mark-by-mark walkthroughs · SAT / AP-Feeder / ON / BC styles练习题配套答案 · 逐分讲解 · SAT / AP 衔接 / 安 / 卑诗省考风格

EASY易 MEDIUM中 HARD难 🇺🇸 US美 🇨🇦 ON安 🇨🇦 BC卑 SAT-style MCQSAT 风格选择题 AP-feeder FRQAP 衔接简答题 ON Provincial-style安大略省考风格 BC Provincial-style卑诗省考风格 Honors荣誉级
PART I  ·  SHORT RESPONSE · SOLUTIONS第一部分  ·  短答题 · 详解SAT MCQ + ON/BC short answer · 22 marksSAT 选择题 + 安/卑省考短答 · 共 22 分

Section A · Worked SolutionsA 部分 · 详细解答

Q1EASY易 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §1 Polynomial Vocabulary多项式术语 · HSA-SSE.A.1 [3 marks][3 分]

Degree, leading coefficient, and constant term of $p(x) = -4x^{5} + 7x^{3} - 2x^{2} + 9$.求 $p(x) = -4x^{5} + 7x^{3} - 2x^{2} + 9$ 的次数、首项系数与常数项。

Answer:答案：  (A)  $5,\ -4,\ 9$

(a) Identify the three structural pieces辨识三个结构部分 M1·A1·A1

Write $p(x)$ in descending powers of $x$ (it already is). The degree is the highest exponent on a non-zero term: $5$. The leading coefficient is the coefficient of that highest-degree term: $-4$. The constant term is the coefficient of $x^{0}$, i.e. the term with no $x$: $9$. So the triple is $(5,\ -4,\ 9)$, matching (A).将 $p(x)$ 按 $x$ 的降幂排列（题目已经排好）。次数是非零项中最高的指数：$5$。首项系数是该最高次项的系数：$-4$。常数项是 $x^{0}$ 的系数，即不含 $x$ 的那一项：$9$。所以三元组为 $(5,\ -4,\ 9)$，对应 (A)。
Why the wrong choices fail.干扰项分析。
  • (B) $5,\ 7,\ 9$, reads degree correctly but picks the coefficient of the second-highest term ($7x^{3}$) as the leading coefficient. Trap: students who skip past the $-4x^{5}$ because the sign is negative.次数读对了，但把次高项（$7x^{3}$）的系数当成首项系数。陷阱：学生因 $-4x^{5}$ 是负号而跳过了它。
  • (C) $4,\ -4,\ -2$, off-by-one on the exponent (degree $4$ instead of $5$) and picks the $-2x^{2}$ constant by mistake; the constant term is the term with no $x$, not the smallest-power term that still has $x$ in it.次数差一（写成 $4$ 而不是 $5$），又错把 $-2x^{2}$ 当成常数项；常数项是不含 $x$ 的那一项，而不是仍带 $x$ 的最低次项。
  • (D) $9,\ -4,\ 5$, swaps degree and constant term: confuses "the constant" ($9$) with "the degree" ($5$).把次数与常数项位置互换：把"常数"（$9$）和"次数"（$5$）搞混。
Standard form first; then read off the three labels in one sweep.先化为标准形式，再一次性读出三个标签。 Polynomial vocabulary questions are 100% reading comprehension after the polynomial is in descending order. The discipline: rewrite $p(x)$ with terms ordered by decreasing degree before answering anything, even if it already looks ordered , this catches polynomials given as $9 - 2x^{2} + 7x^{3} - 4x^{5}$ where the leading coefficient hides at the right end. The "constant term" is the value $p(0)$; in standard form it's literally the term with no $x$. This three-token sweep (degree, leading coefficient, constant) is the muscle memory for §3.5 end-behaviour (degree + sign of lead) and §3.4 rational root theorem (constant + lead).多项式术语题，只要把多项式按降幂排好，剩下的百分之百是阅读理解。规矩：动笔前先把 $p(x)$ 按降幂重写一遍，哪怕看起来已经是降幂——这能识别像 $9 - 2x^{2} + 7x^{3} - 4x^{5}$ 这种把首项系数藏在右端的写法。"常数项"等于 $p(0)$，在标准形式下就是不含 $x$ 的那一项。这套三步扫描（次数、首项系数、常数项）将成为 §3.5 末端行为（次数 + 首项符号）和 §3.4 有理根定理（常数项 + 首项系数）的肌肉记忆。
Q2EASY易 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §3 Remainder Theorem余数定理 · HSA-SSE.A.2 [3 marks][3 分]

Remainder when $p(x) = x^{3} - 4x^{2} + 5x - 7$ is divided by $(x - 2)$.求 $p(x) = x^{3} - 4x^{2} + 5x - 7$ 被 $(x - 2)$ 除时的余数。

Answer:答案：  (A)  $-5$

(a) Apply the remainder theorem应用余数定理 M1·A1

The remainder theorem: when $p(x)$ is divided by $(x - c)$, the remainder is $p(c)$. Here $c = 2$, so the remainder is $p(2)$.余数定理：$p(x)$ 被 $(x - c)$ 除时余数为 $p(c)$。此处 $c = 2$，故余数为 $p(2)$。

(b) Evaluate $p(2)$计算 $p(2)$ A1

$$ p(2) \;=\; 2^{3} - 4(2)^{2} + 5(2) - 7 \;=\; 8 - 16 + 10 - 7 \;=\; -5. $$ Remainder is $-5$, matching (A).余数为 $-5$，对应 (A)。
Why the wrong choices fail.干扰项分析。
  • (B) $-3$, sign error inside $-4(2)^{2}$: student computes $-4 \cdot 4 = -8$ instead of $-16$ (or applies $-4 \cdot 2^{2} = (-4 \cdot 2)^{2}$, an order-of-operations slip).在 $-4(2)^{2}$ 一项出现符号或运算错误：把 $-4 \cdot 4$ 算成 $-8$ 而非 $-16$（或把 $-4 \cdot 2^{2}$ 当成 $(-4 \cdot 2)^{2}$，运算次序失误）。
  • (C) $0$, the "of course it's a factor" trap; students who memorise that easy MCQ remainders are zero. But the remainder theorem must be computed, not guessed."它当然是因式"型陷阱；学生死记简单选择题的余数都是 $0$。但余数定理必须算出，不能猜。
  • (D) $5$, drops the overall negative sign at the end, or evaluates $p(-2)$ instead of $p(2)$ by misreading the divisor sign.最后丢掉了整体负号，或者把除式符号看反，代入 $p(-2)$ 而非 $p(2)$。
The remainder theorem turns long division into one substitution.余数定理把长除法压成一次代入。 If the divisor is linear of the form $(x - c)$, never carry out polynomial long division for the remainder , just compute $p(c)$. The sign trap is loud: divisor $(x - 2)$ means $c = +2$, but divisor $(x + 2)$ means $c = -2$ (because $x + 2 = x - (-2)$). The remainder theorem is the engine behind the factor theorem (§3.3): $p(c) = 0 \Leftrightarrow (x - c)$ is a factor. Internalise the substitution-instead-of-division move now; it pays for itself in §3.4 (rational root theorem), Q6, Q12.如果除式是 $(x - c)$ 这种一次形式，求余数千万不要做多项式长除法，直接算 $p(c)$ 即可。符号陷阱很明显：除式 $(x - 2)$ 对应 $c = +2$；除式 $(x + 2)$ 对应 $c = -2$（因为 $x + 2 = x - (-2)$）。余数定理也是因式定理（§3.3）的底层引擎：$p(c) = 0 \Leftrightarrow (x - c)$ 是因式。现在就把"用代入代替除法"内化下来，在 §3.4（有理根定理）、Q6、Q12 中会反复见到回报。
Q3MEDIUM中 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §5 End Behaviour末端行为 · HSF-IF.B.4 [4 marks][4 分]

End behaviour of $f(x) = -2x^{4} + 6x^{3} - x + 8$ as $x \to \pm \infty$.$x \to \pm \infty$ 时 $f(x) = -2x^{4} + 6x^{3} - x + 8$ 的末端行为。

Answer:答案：  (B)  $f(x) \to -\infty$ on both ends两端均 $f(x) \to -\infty$

(a) Isolate the leading term分离首项 M1·A1

For large $|x|$, every lower-degree term is dominated by the leading term, so $f(x) \approx -2x^{4}$. Read off two pieces from $-2x^{4}$: the degree $n = 4$ (even) and the sign of the leading coefficient ($-2 < 0$, i.e. negative).当 $|x|$ 很大时，每个低次项都被首项压制，因此 $f(x) \approx -2x^{4}$。从 $-2x^{4}$ 读出两点：次数 $n = 4$（偶数），首项系数符号为负（$-2 < 0$）。

(b) Combine degree-parity and sign把次数奇偶与符号合起来看 A1·A1

Even degree means both ends behave the same way (left and right go to the same infinity). Negative leading coefficient flips them down. So $f(x) \to -\infty$ as $x \to +\infty$ and as $x \to -\infty$. Matches (B).偶次意味着两端方向一致（左右趋向同一无穷）。首项系数为负把方向翻下。故 $x \to +\infty$ 与 $x \to -\infty$ 时均有 $f(x) \to -\infty$，对应 (B)。
Why the wrong choices fail.干扰项分析。
  • (A) $f(x) \to +\infty$ on both ends两端均 $f(x) \to +\infty$, would be correct for $+2x^{4}$ (positive lead, even degree, both ends up); ignores the sign of the leading coefficient.这是 $+2x^{4}$ 的答案（正首项 + 偶次，两端都向上）；忽略了首项系数的符号。
  • (C) opposite ends, with $+\infty$ on the left两端方向相反，左端 $+\infty$, would be correct for an odd-degree polynomial with negative lead (e.g. $-2x^{5}$); misreads the parity of the degree.这是奇次负首项多项式（如 $-2x^{5}$）的答案；把次数奇偶判错了。
  • (D) opposite ends, with $+\infty$ on the right两端方向相反，右端 $+\infty$, would be correct for an odd-degree polynomial with positive lead (e.g. $+2x^{5}$); double error on degree-parity and sign.这是奇次正首项多项式（如 $+2x^{5}$）的答案；奇偶与符号双双判错。
Even/odd $\to$ same/opposite; sign of lead $\to$ up/down.偶 / 奇 $\to$ 同 / 异；首项符号 $\to$ 上 / 下。 The end-behaviour mnemonic is two binary toggles: parity of degree decides whether the two ends agree (even = same direction) or disagree (odd = opposite directions), and sign of leading coefficient decides whether the right end goes up ($+$) or down ($-$). Build the table in your head in four seconds: $(+,$ even$) \to$ up/up; $(-,$ even$) \to$ down/down; $(+,$ odd$) \to$ down/up; $(-,$ odd$) \to$ up/down. Lower-degree terms (the $+6x^{3} - x + 8$ here) are irrelevant for end behaviour , they only shift the middle of the graph, not the wings. SAT MCQ writers love seeding the answer set with all four parity/sign combinations.末端行为口诀是两组二选一开关：次数奇偶决定两端方向是同（偶 = 同向）还是异（奇 = 反向）；首项系数符号决定右端朝上（$+$）或朝下（$-$）。四秒内在脑中建表：$(+,$ 偶$) \to$ 上 / 上；$(-,$ 偶$) \to$ 下 / 下；$(+,$ 奇$) \to$ 下 / 上；$(-,$ 奇$) \to$ 上 / 下。低次项（此处的 $+6x^{3} - x + 8$）对末端行为无影响——只挪动图象中段，不动两翼。SAT 选择题作者最爱把四种奇偶 / 符号组合一锅端给你。
Q4MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §2 Polynomial Long Division多项式长除法 · MHF4U Polynomial & Rational FunctionsMHF4U 多项式与有理函数 [6 marks][6 分]

Divide $p(x) = 2x^{3} + 5x^{2} - 4x + 3$ by $d(x) = x + 3$ using polynomial long division. (a) Tableau, $q(x)$, $r$. (b) Division identity.用多项式长除法将 $p(x) = 2x^{3} + 5x^{2} - 4x + 3$ 除以 $d(x) = x + 3$。(a) 竖式、$q(x)$、$r$。(b) 除法恒等式。

Answer:答案：  (a) $q(x) = 2x^{2} - x - 1$, $r = 6$  ·  (b) $2x^{3} + 5x^{2} - 4x + 3 = (x + 3)(2x^{2} - x - 1) + 6$

(a) Long-division tableau长除法竖式 M1·M1·A1·A1

Divide $2x^{3} + 5x^{2} - 4x + 3$ by $x + 3$ step by step. At each step, divide the current leading term by $x$ to get the next quotient term, then multiply and subtract.将 $2x^{3} + 5x^{2} - 4x + 3$ 逐步除以 $x + 3$。每一步把当前首项除以 $x$ 得到下一个商项，再相乘并相减。
  • Step 1.第 1 步。 $2x^{3} \div x = 2x^{2}$. Multiply: $2x^{2}(x + 3) = 2x^{3} + 6x^{2}$. Subtract from $2x^{3} + 5x^{2}$: $(2x^{3} + 5x^{2}) - (2x^{3} + 6x^{2}) = -x^{2}$. Bring down $-4x$: new partial dividend $-x^{2} - 4x$.$2x^{3} \div x = 2x^{2}$。乘：$2x^{2}(x + 3) = 2x^{3} + 6x^{2}$。从 $2x^{3} + 5x^{2}$ 减：$(2x^{3} + 5x^{2}) - (2x^{3} + 6x^{2}) = -x^{2}$。把 $-4x$ 拉下：新被除项为 $-x^{2} - 4x$。
  • Step 2.第 2 步。 $-x^{2} \div x = -x$. Multiply: $-x(x + 3) = -x^{2} - 3x$. Subtract: $(-x^{2} - 4x) - (-x^{2} - 3x) = -x$. Bring down $+3$: new partial dividend $-x + 3$.$-x^{2} \div x = -x$。乘：$-x(x + 3) = -x^{2} - 3x$。减：$(-x^{2} - 4x) - (-x^{2} - 3x) = -x$。把 $+3$ 拉下：新被除项为 $-x + 3$。
  • Step 3.第 3 步。 $-x \div x = -1$. Multiply: $-1(x + 3) = -x - 3$. Subtract: $(-x + 3) - (-x - 3) = 6$. Degree of remainder $(0) <$ degree of divisor $(1)$, so stop.$-x \div x = -1$。乘：$-1(x + 3) = -x - 3$。减：$(-x + 3) - (-x - 3) = 6$。余数次数 $(0) <$ 除式次数 $(1)$，过程结束。
Quotient $q(x) = 2x^{2} - x - 1$. Remainder $r = 6$.商 $q(x) = 2x^{2} - x - 1$，余数 $r = 6$。

(b) Division identity除法恒等式 M1·A1

The identity $p(x) = d(x) \cdot q(x) + r$ with values substituted:将数值代入恒等式 $p(x) = d(x) \cdot q(x) + r$： $$ 2x^{3} + 5x^{2} - 4x + 3 \;=\; (x + 3)(2x^{2} - x - 1) + 6. $$ Cross-check via remainder theorem:用余数定理交叉验证： dividing by $(x + 3)$ should give remainder $p(-3) = 2(-27) + 5(9) - 4(-3) + 3 = -54 + 45 + 12 + 3 = 6$. $\checkmark$被 $(x + 3)$ 除时余数应等于 $p(-3) = 2(-27) + 5(9) - 4(-3) + 3 = -54 + 45 + 12 + 3 = 6$。$\checkmark$
"Divide-multiply-subtract-bring down", and keep place-value columns aligned."除—乘—减—拉下"，并保持位值列对齐。 The single most common ON provincial error is a missing-term column: writing $2x^{3} + 5x^{2} - 4x + 3$ without explicitly leaving a column for any missing power slot. Here every power is present, but for $x^{3} + 1 \div (x + 1)$ you must write $x^{3} + 0 x^{2} + 0 x + 1$ before dividing , otherwise the subtractions misalign and the remainder is wrong. Sign-subtract trap (Step 2 here): subtracting $-3x$ from $-4x$ is $-4x - (-3x) = -4x + 3x = -x$, not $-7x$. The remainder-theorem cross-check at the end is free insurance , if $p(-3) \ne r$, the long division has an error somewhere.安大略省考最常见的错就是缺项列：写 $2x^{3} + 5x^{2} - 4x + 3$ 时没有为缺失的幂留出列。此题每个幂都齐，但 $x^{3} + 1 \div (x + 1)$ 必须先写成 $x^{3} + 0 x^{2} + 0 x + 1$ 再除——否则减法错位、余数算错。第 2 步的符号陷阱：用 $-4x$ 减 $-3x$ 是 $-4x - (-3x) = -4x + 3x = -x$，不是 $-7x$。最后用余数定理交叉检验是免费的保险：若 $p(-3) \ne r$，长除法肯定有错。
Q5EASY易 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §2 Synthetic Division综合除法 · BC PC12 polynomial functions and equations卑诗 PC12 多项式函数与方程 [6 marks][6 分]

$p(x) = x^{3} - 6x^{2} + 11x - 6$. (a) Synthetic division by $(x - 1)$ with tableau. (b) $q(x)$ and $r$. (c) Factor theorem conclusion.$p(x) = x^{3} - 6x^{2} + 11x - 6$。(a) 用 $(x - 1)$ 做综合除法并给出竖式。(b) $q(x)$ 与 $r$。(c) 因式定理结论。

Answer:答案：  (a) tableau below竖式见下  ·  (b) $q(x) = x^{2} - 5x + 6$, $r = 0$  ·  (c) $(x - 1)$ is a factor by the factor theorem由因式定理，$(x - 1)$ 是因式

(a) Synthetic-division tableau综合除法竖式 M1·A1·A1

The divisor is $(x - 1)$, so the synthetic-division root is $c = 1$. Write the coefficients of $p(x)$ in descending order: $1,\ -6,\ 11,\ -6$. Bring down the leading $1$; then multiply by $c = 1$ and add to the next column. Repeat.除式为 $(x - 1)$，故综合除法的根 $c = 1$。把 $p(x)$ 的系数按降幂写出：$1,\ -6,\ 11,\ -6$。先把首项 $1$ 拉下，再乘以 $c = 1$ 并加到下一列，重复此过程。
  • Bring down $1$.把 $1$ 拉下。
  • $1 \cdot 1 = 1$; $-6 + 1 = -5$.
  • $-5 \cdot 1 = -5$; $11 + (-5) = 6$.
  • $6 \cdot 1 = 6$; $-6 + 6 = 0$.
Bottom row: $1,\ -5,\ 6 \mid 0$. The first three numbers are the coefficients of the quotient (degree drops by 1), and the boxed $0$ is the remainder.底行：$1,\ -5,\ 6 \mid 0$。前三个数是商的系数（次数降 1），被框出的 $0$ 是余数。

(b) Read off $q(x)$ and $r$读出 $q(x)$ 与 $r$ A1·A1

Quotient: $q(x) = x^{2} - 5x + 6$. Remainder: $r = 0$.商 $q(x) = x^{2} - 5x + 6$，余数 $r = 0$。

(c) Factor-theorem conclusion因式定理结论 R1

Since the remainder $p(1) = 0$, by the factor theorem $(x - 1)$ is a factor of $p(x)$.由于余数 $p(1) = 0$，根据因式定理，$(x - 1)$ 是 $p(x)$ 的因式。
Synthetic division is long division with the column-bookkeeping removed.综合除法就是把列对齐这步省掉的长除法。 It works only when the divisor is linear $(x - c)$ , for higher-degree divisors you must use long division. Two locked-in habits: (1) write the coefficients in descending order with zeros for any missing power (e.g. $x^{3} + 1 \to$ coefficients $1, 0, 0, 1$), and (2) the sign of $c$ flips: divisor $(x - 1) \to c = +1$; divisor $(x + 1) \to c = -1$. The bottom row's last entry is the remainder; the rest are quotient coefficients of one degree lower than the dividend. Bonus: this question's $q(x) = x^{2} - 5x + 6 = (x - 2)(x - 3)$ factors further, so the full factorization is $p(x) = (x - 1)(x - 2)(x - 3)$ , zeros at $1, 2, 3$, a tidy classroom example. Reuse this pattern as a memory hook.综合除法仅当除式为一次 $(x - c)$ 时适用——除式次数更高就必须用长除法。两条必须记住的习惯：(1) 系数按降幂写出，缺项的幂补 0（如 $x^{3} + 1$ 的系数为 $1, 0, 0, 1$）；(2) $c$ 的符号要翻：除式 $(x - 1) \to c = +1$，除式 $(x + 1) \to c = -1$。底行最后一项是余数，其余是比被除式低一次的商的系数。附加：本题 $q(x) = x^{2} - 5x + 6 = (x - 2)(x - 3)$ 还可继续分解，所以完全因式分解是 $p(x) = (x - 1)(x - 2)(x - 3)$，零点为 $1, 2, 3$——课本级整洁示例，可作记忆挂钩反复使用。
PART II  ·  EXTENDED RESPONSE · SOLUTIONS第二部分  ·  简答题 · 详解AP-feeder FRQ + honors · 36 marksAP 衔接简答题 + 荣誉级 · 共 36 分

Section B · Worked SolutionsB 部分 · 详细解答

Q6MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §3 Factor Theorem因式定理 · HSF-IF.C.8 [8 marks][8 分]

$p(x) = x^{3} + 2x^{2} - 5x - 6$. (a) $p(-1)$, $p(2)$, $p(-3)$. (b) Factor-theorem identification. (c) Complete factorisation and zeros.$p(x) = x^{3} + 2x^{2} - 5x - 6$。(a) 求 $p(-1)$、$p(2)$、$p(-3)$。(b) 用因式定理辨识。(c) 完全因式分解与零点。

Answer:答案：  (a) $p(-1) = 0$, $p(2) = 0$, $p(-3) = 0$  ·  (b) all three are factors三个都是因式  ·  (c) $p(x) = (x + 1)(x - 2)(x + 3)$; zeros $-1, 2, -3$零点 $-1, 2, -3$

(a) Evaluate at the three candidates在三个候选根处求值 M1·A1·A1

$$ p(-1) \;=\; (-1)^{3} + 2(-1)^{2} - 5(-1) - 6 \;=\; -1 + 2 + 5 - 6 \;=\; 0. $$ $$ p(2) \;=\; (2)^{3} + 2(2)^{2} - 5(2) - 6 \;=\; 8 + 8 - 10 - 6 \;=\; 0. $$ $$ p(-3) \;=\; (-3)^{3} + 2(-3)^{2} - 5(-3) - 6 \;=\; -27 + 18 + 15 - 6 \;=\; 0. $$

(b) Factor-theorem identification因式定理辨识 R1·A1

The factor theorem states: $(x - c)$ is a factor of $p(x)$ iff $p(c) = 0$. Each evaluation in (a) gave zero, so all three of $(x + 1)$, $(x - 2)$, $(x + 3)$ are factors of $p(x)$.因式定理：$(x - c)$ 是 $p(x)$ 的因式当且仅当 $p(c) = 0$。(a) 中三次求值都得 $0$，故 $(x + 1)$、$(x - 2)$、$(x + 3)$ 都是 $p(x)$ 的因式。

(c) Complete factorisation and zeros完全因式分解与零点 M1·A1·A1

A cubic has at most three real zeros and at most three linear factors. We have found three distinct factors, so they exhaust $p$ up to a constant. The leading coefficient of the product $(x + 1)(x - 2)(x + 3)$ expanded is $1$, matching the leading $1$ of $p(x)$, so no extra scalar is needed:三次多项式至多有三个实零点、至多三个一次因式。我们已找到三个不同因式，故除常数倍外已穷尽 $p$。$(x + 1)(x - 2)(x + 3)$ 展开后的首项系数为 $1$，与 $p(x)$ 的首项 $1$ 相符，无需额外标量： $$ p(x) \;=\; (x + 1)(x - 2)(x + 3). $$ Verify by expansion.展开验证。 $(x + 1)(x + 3) = x^{2} + 4x + 3$, then，再 $(x^{2} + 4x + 3)(x - 2) = x^{3} - 2x^{2} + 4x^{2} - 8x + 3x - 6 = x^{3} + 2x^{2} - 5x - 6$. $\checkmark$ The three real zeros of $p$ are therefore $x = -1$, $x = 2$, $x = -3$.因此 $p$ 的三个实零点为 $x = -1$、$x = 2$、$x = -3$。
Factor theorem turns the cubic into three independent yes/no checks.因式定理把三次问题压成三次独立的是 / 否检验。 Once you have a candidate root list, evaluation is the fastest test , one substitution per candidate, no division required. AP graders explicitly award the R1 for citing the factor theorem by name; writing "since $p(-1) = 0$, $(x + 1)$ is a factor" without the citation loses a mark on HSF-IF.C.8. The leading-coefficient check at the end is the audit: if the product of factors has lead $1$ but $p$ has lead $a \ne 1$, the complete factorisation is $p(x) = a \cdot \prod (x - c_{i})$ , do not omit the scalar. This three-zero cubic is the prototype for Q12's sign-chart machinery.有候选根表后，代入求值是最快的检验：每个候选只需一次代入，无需做除法。AP 阅卷会明确奖励显式引用因式定理的 R1 分；只写"因为 $p(-1) = 0$，所以 $(x + 1)$ 是因式"而不引用定理，会在 HSF-IF.C.8 上丢分。最后的首项系数检查是审计：若因式乘积首项为 $1$，但 $p$ 首项为 $a \ne 1$，那么完全因式分解为 $p(x) = a \cdot \prod (x - c_{i})$——不要漏掉这个标量。本题的三零点三次式是 Q12 符号表机制的原型。
Q7HARD难 🇨🇦 ON安 ON Provincial-style安大略省考风格 §4 Rational Root Theorem有理根定理 · MHF4U Polynomial & Rational FunctionsMHF4U 多项式与有理函数 [10 marks][10 分]

$2x^{3} - 3x^{2} - 11x + 6 = 0$. (a) Candidate rational roots. (b) Find one rational root. (c) Depress and factor the quadratic. (d) All three roots and full factorisation.$2x^{3} - 3x^{2} - 11x + 6 = 0$。(a) 候选有理根。(b) 找出一个有理根。(c) 降次并因式分解所得二次式。(d) 三个根与完全因式分解。

Answer:答案：  (a) $\pm \{1, 2, 3, 6, \tfrac{1}{2}, \tfrac{3}{2}\}$  ·  (b) $x = 3$  ·  (c) depressed quadratic降次二次式 $2x^{2} + 3x - 2 = (2x - 1)(x + 2)$  ·  (d) roots根 $3,\ \tfrac{1}{2},\ -2$; full factorisation完全因式分解 $(x - 3)(2x - 1)(x + 2)$

(a) Candidate $\pm p/q$ list候选 $\pm p/q$ 清单 M1·A1·A1

The rational root theorem: any rational root $p/q$ in lowest terms must have $p$ dividing the constant term and $q$ dividing the leading coefficient. Here constant $= 6$, lead $= 2$.有理根定理：任意最简形式的有理根 $p/q$ 必须满足 $p$ 整除常数项、$q$ 整除首项系数。此处常数项 $= 6$，首项系数 $= 2$。
Divisors of $6$ (the $p$-values): $1, 2, 3, 6$.$6$ 的因数（$p$ 的取值）：$1, 2, 3, 6$。
Divisors of $2$ (the $q$-values): $1, 2$.$2$ 的因数（$q$ 的取值）：$1, 2$。
All $\pm p/q$: $\pm \{1, 2, 3, 6\} \cup \pm \{\tfrac{1}{2}, \tfrac{2}{2}, \tfrac{3}{2}, \tfrac{6}{2}\} = \pm \{1, 2, 3, 6, \tfrac{1}{2}, \tfrac{3}{2}\}$ after dropping duplicates ($\tfrac{2}{2} = 1$, $\tfrac{6}{2} = 3$). Twelve candidates in all.所有 $\pm p/q$：$\pm \{1, 2, 3, 6\} \cup \pm \{\tfrac{1}{2}, \tfrac{2}{2}, \tfrac{3}{2}, \tfrac{6}{2}\} = \pm \{1, 2, 3, 6, \tfrac{1}{2}, \tfrac{3}{2}\}$（去重 $\tfrac{2}{2} = 1$、$\tfrac{6}{2} = 3$ 后）。共 12 个候选。

(b) Test candidates by substitution用代入法检验候选 M1·A1

Test $x = 1$: $2 - 3 - 11 + 6 = -6 \ne 0$. Test $x = -1$: $-2 - 3 + 11 + 6 = 12 \ne 0$. Test $x = 2$: $16 - 12 - 22 + 6 = -12 \ne 0$. Test $x = 3$: $2(27) - 3(9) - 11(3) + 6 = 54 - 27 - 33 + 6 = 0$. $\checkmark$试 $x = 1$：$2 - 3 - 11 + 6 = -6 \ne 0$。试 $x = -1$：$-2 - 3 + 11 + 6 = 12 \ne 0$。试 $x = 2$：$16 - 12 - 22 + 6 = -12 \ne 0$。试 $x = 3$：$2(27) - 3(9) - 11(3) + 6 = 54 - 27 - 33 + 6 = 0$。$\checkmark$
Rational root: $x = 3$.有理根：$x = 3$。

(c) Depress the cubic via synthetic division用综合除法降次 M1·A1·A1

Synthetic division by $(x - 3)$ on coefficients $2, -3, -11, 6$:对系数 $2, -3, -11, 6$ 用 $(x - 3)$ 做综合除法：
  • Bring down $2$. Multiply $2 \cdot 3 = 6$; $-3 + 6 = 3$.把 $2$ 拉下。乘 $2 \cdot 3 = 6$；$-3 + 6 = 3$。
  • Multiply $3 \cdot 3 = 9$; $-11 + 9 = -2$.乘 $3 \cdot 3 = 9$；$-11 + 9 = -2$。
  • Multiply $-2 \cdot 3 = -6$; $6 + (-6) = 0$.乘 $-2 \cdot 3 = -6$；$6 + (-6) = 0$。
Quotient: $2x^{2} + 3x - 2$, remainder $0$. Factor the quadratic by AC-method: $AC = 2 \cdot (-2) = -4$; integers with product $-4$ and sum $3$ are $4$ and $-1$. Split and group:商：$2x^{2} + 3x - 2$，余数 $0$。用 AC 法分解此二次式：$AC = 2 \cdot (-2) = -4$；积为 $-4$、和为 $3$ 的整数是 $4$ 和 $-1$。拆项分组： $$ 2x^{2} + 4x - x - 2 \;=\; 2x(x + 2) - 1(x + 2) \;=\; (2x - 1)(x + 2). $$

(d) All three roots and full factorisation三个根与完全因式分解 A1·A1

Roots: $x = 3$, $2x - 1 = 0 \Rightarrow x = \tfrac{1}{2}$, $x + 2 = 0 \Rightarrow x = -2$.根：$x = 3$、$2x - 1 = 0 \Rightarrow x = \tfrac{1}{2}$、$x + 2 = 0 \Rightarrow x = -2$。
Full factorisation:完全因式分解： $$ 2x^{3} - 3x^{2} - 11x + 6 \;=\; (x - 3)(2x - 1)(x + 2). $$ Audit by expansion of leading and constant terms.用首项与常数项审核。 Leading: $1 \cdot 2 \cdot 1 = 2$ $\checkmark$. Constant: $(-3)(-1)(2) = 6$ $\checkmark$.首项：$1 \cdot 2 \cdot 1 = 2$ $\checkmark$。常数项：$(-3)(-1)(2) = 6$ $\checkmark$。
RRT shrinks an infinite search to a finite list , but only fractions in lowest terms.有理根定理把无限搜索压成有限清单——但仅限最简分数。 The classic trap: students with leading coefficient $\ne 1$ forget to include the $\pm p/q$ fractions and only test integer candidates , missing roots like $\tfrac{1}{2}$ here. The second trap: testing every candidate when an audit shortcut exists. The sum of all roots equals $-b/a = 3/2$ (Vieta); once you find $x = 3$, the remaining two roots sum to $\tfrac{3}{2} - 3 = -\tfrac{3}{2}$, and their product equals (last coefficient of depressed quadratic) $\div$ (lead of depressed) $= -2/2 = -1$. Then $x = \tfrac{1}{2}$ and $x = -2$ are forced. Reading roots from coefficients via Vieta is the speed-multiplier MHF4U markers reward. Always cite the rational root theorem by name to bank the R1.经典陷阱：首项系数 $\ne 1$ 时，学生忘了纳入 $\pm p/q$ 分数候选，只测整数——会漏掉本题这种 $\tfrac{1}{2}$。第二个陷阱：明明有速算捷径却把每个候选都测一遍。由韦达定理，所有根之和 $= -b/a = 3/2$；一旦找到 $x = 3$，另两根之和 $= \tfrac{3}{2} - 3 = -\tfrac{3}{2}$，积 $=$ 降次二次式的常数项 $\div$ 首项系数 $= -2/2 = -1$。则 $x = \tfrac{1}{2}$ 与 $x = -2$ 被锁定。用韦达从系数读出根是 MHF4U 阅卷会奖励的速算思路。务必在答题时显式引用"有理根定理"以拿稳 R1。
Q8MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §5 Zeros, Multiplicity & End Behaviour零点、重数与末端行为 · HSF-IF.C.7 [9 marks][9 分]

$f(x) = -(x + 2)^{2}(x - 1)(x - 3)$. (a) Degree, sign of lead, end behaviour. (b) $x$-intercepts with multiplicities & cross/touch. (c) $y$-intercept. (d) Sketch.$f(x) = -(x + 2)^{2}(x - 1)(x - 3)$。(a) 次数、首项符号、末端行为。(b) $x$ 轴截距、重数与穿过 / 相切。(c) $y$ 轴截距。(d) 作图。

Answer:答案：  (a) degree $4$, lead $-1 < 0$, both ends $\to -\infty$次数 $4$，首项 $-1 < 0$，两端均 $\to -\infty$  ·  (b) $x = -2$ (mult 2, touch)（重数 2，相切）; $x = 1$ (mult 1, cross)（重数 1，穿过）; $x = 3$ (mult 1, cross)（重数 1，穿过）  ·  (c) $(0, -12)$  ·  (d) sketch description below作图说明见下

(a) Degree, lead sign, end behaviour次数、首项符号、末端行为 M1·A1

Degree = sum of multiplicities = $2 + 1 + 1 = 4$ (even). Leading coefficient of the expanded form: multiply the $x$-coefficient of each factor and the overall $-1$: $(-1)(1)(1)(1) \cdot 1 = -1$ (negative). With even degree and negative lead, end behaviour: $f(x) \to -\infty$ as $x \to +\infty$ and as $x \to -\infty$ (both ends down).次数 = 各重数之和 = $2 + 1 + 1 = 4$（偶数）。展开后的首项系数：把每个因式中 $x$ 的系数与整体 $-1$ 相乘：$(-1)(1)(1)(1) \cdot 1 = -1$（负）。偶次 + 负首项，末端行为：$x \to +\infty$ 与 $x \to -\infty$ 时均有 $f(x) \to -\infty$（两端朝下）。

(b) $x$-intercepts with multiplicities$x$ 轴截距与重数 A1·A1·A1

Read directly from the factored form. Each factor $(x - c)^{m}$ contributes a zero $x = c$ of multiplicity $m$:直接从因式形式读出。每个因式 $(x - c)^{m}$ 提供一个重数为 $m$ 的零点 $x = c$：
  • $x = -2$, multiplicity $2$ (even) $\to$ graph touches the $x$-axis at $(-2, 0)$ and bounces back.$x = -2$，重数 $2$（偶）$\to$ 图象在 $(-2, 0)$ 处相切于 $x$ 轴并反弹。
  • $x = 1$, multiplicity $1$ (odd) $\to$ graph crosses through $(1, 0)$.$x = 1$，重数 $1$（奇）$\to$ 图象穿过 $(1, 0)$。
  • $x = 3$, multiplicity $1$ (odd) $\to$ graph crosses through $(3, 0)$.$x = 3$，重数 $1$（奇）$\to$ 图象穿过 $(3, 0)$。

(c) $y$-intercept$y$ 轴截距 A1

$$ f(0) \;=\; -(0 + 2)^{2}(0 - 1)(0 - 3) \;=\; -(4)(-1)(-3) \;=\; -12. $$ So the $y$-intercept is $(0, -12)$.故 $y$ 轴截距为 $(0, -12)$。

(d) Sketch and verification of sign on the four intervals作图并验证四个区间上的符号 M1·A1

Zeros at $-2, 1, 3$ split the $x$-axis into four intervals; combined with end behaviour and multiplicity, the sign of $f$ on each is forced. Check $x = 0$: $f(0) = -12 < 0$ on $(-2, 1)$. By the multiplicity rules (sign flips at odd-mult zeros, stays at even-mult zeros), the sign pattern from left to right is:零点 $-2, 1, 3$ 把 $x$ 轴分成四段；结合末端行为与重数，每段上 $f$ 的符号被锁定。检验 $x = 0$：$f(0) = -12 < 0$，即在 $(-2, 1)$ 上为负。由重数法则（奇重数零点处符号翻转，偶重数零点处符号保持），自左向右的符号模式为：
$(-\infty, -2)$: $\;-\;$; bounce at $x = -2$ (no sign change)$x = -2$ 处反弹（不变号） $\to$ $(-2, 1)$: $\;-\;$; cross at $x = 1$$x = 1$ 处穿过 $\to$ $(1, 3)$: $\;+\;$; cross at $x = 3$$x = 3$ 处穿过 $\to$ $(3, \infty)$: $\;-\;$.
Sketch: starts at $-\infty$ on the left, rises toward $x = -2$, just touches the axis there and falls back, deep negative dip through $(0, -12)$, crosses up through $(1, 0)$, peaks in $(1, 3)$, crosses back down through $(3, 0)$, and falls to $-\infty$ on the right.作图：左端自 $-\infty$ 升起，靠近 $x = -2$ 时刚好相切又落回，经 $(0, -12)$ 形成深谷，在 $(1, 0)$ 处向上穿过，在 $(1, 3)$ 间到达局部峰值，在 $(3, 0)$ 处再向下穿过，右端落向 $-\infty$。
Multiplicity is the sign-change toggle.重数是变号开关。 At a zero of odd multiplicity, $f$ changes sign (graph crosses the $x$-axis , multiplicity $1$ = straight crossing, multiplicity $3$ = inflection-shaped crossing). At a zero of even multiplicity, $f$ keeps its sign (graph touches the axis and bounces back , multiplicity $2$ = parabolic kiss, multiplicity $4$ = flatter kiss). Once one interval's sign is known (use $f(0)$ , the $y$-intercept is free), the rest are forced by walking left/right and flipping signs only at odd-multiplicity zeros. AP graders mark all three sub-rubrics: degree+sign $\to$ end behaviour (the M1), multiplicity $\to$ cross/touch (the A1s), and the actual sketch must show the correct turning behaviour at each zero, not just hit the points. A "touch and bounce" at $x = -2$ that looks like a "cross" loses an A1.在奇重数零点处 $f$ 变号（图象穿过 $x$ 轴：重数 $1$ 是直穿，重数 $3$ 是带拐点的穿过）。在偶重数零点处 $f$ 保号（相切并反弹：重数 $2$ 是抛物形相吻，重数 $4$ 更平缓）。只要某一区间的符号已知（用 $f(0)$——$y$ 轴截距是免费数据），其余区间的符号通过向左 / 向右走、只在奇重数零点处翻转就被锁定。AP 阅卷会按三条子标准给分：次数 + 符号 $\to$ 末端行为（M1）、重数 $\to$ 穿过 / 相切（多个 A1），实际作图还必须在每个零点处呈现正确的转折——只是点连过去是不够的。如果在 $x = -2$ 画成"穿过"而非"相切反弹"，就会丢一个 A1。
Q9HARD难Honors荣誉级 🇺🇸 US美 🇨🇦 BC卑 AP-feeder FRQAP 衔接简答题 §6 Complex Conjugate Roots复共轭根 · HSN-CN.A.7 / BC PC12 [9 marks][9 分]

Real cubic, lead $1$, two known roots $x = 3$ and $x = 1 + 2i$. (a) Third root. (b) Conjugate pair $\to$ real quadratic factor $x^{2} - 2x + 5$. (c) Standard form. (d) Geometric meaning.实系数三次式，首项 $1$，已知两根 $x = 3$ 与 $x = 1 + 2i$。(a) 第三个根。(b) 共轭对 $\to$ 实二次因式 $x^{2} - 2x + 5$。(c) 标准形式。(d) 几何意义。

Answer:答案：  (a) third root $x = 1 - 2i$第三个根 $x = 1 - 2i$  ·  (b) $(x - (1 + 2i))(x - (1 - 2i)) = x^{2} - 2x + 5$  ·  (c) $p(x) = x^{3} - 5x^{2} + 11x - 15$  ·  (d) one real $x$-intercept at $x = 3$; irreducible quadratic has no real zeros仅有一个实 $x$ 轴截距 $x = 3$；不可约二次因式无实零点

(a) Third root by complex conjugate root theorem由复共轭根定理求第三个根 R1·A1

The complex conjugate root theorem (HSN-CN.A.7 in US, equivalent under BC PC12 polynomial-equations content) states: if a polynomial has real coefficients and a non-real complex root $a + bi$, then its conjugate $a - bi$ is also a root. Here $p$ has integer (hence real) coefficients and $1 + 2i$ is a root, so $\overline{1 + 2i} = 1 - 2i$ must also be a root. The three roots of the cubic are therefore $3,\ 1 + 2i,\ 1 - 2i$, i.e. third root $x = 1 - 2i$.复共轭根定理（美国 HSN-CN.A.7，卑诗 PC12 多项式方程内容下等价）：若一多项式系数为实数且有非实复根 $a + bi$，则其共轭 $a - bi$ 也是根。此处 $p$ 系数为整数（因而是实数），$1 + 2i$ 是根，故 $\overline{1 + 2i} = 1 - 2i$ 也必为根。三次式的三个根为 $3,\ 1 + 2i,\ 1 - 2i$，即第三个根 $x = 1 - 2i$。

(b) Conjugate pair multiplies to a real quadratic共轭对相乘得到实二次式 M1·A1·A1

Use the difference-of-squares pattern $(A - B)(A + B) = A^{2} - B^{2}$ with $A = x - 1$ and $B = 2i$:取 $A = x - 1$、$B = 2i$，套用平方差公式 $(A - B)(A + B) = A^{2} - B^{2}$： $$ (x - (1 + 2i))(x - (1 - 2i)) \;=\; ((x - 1) - 2i)((x - 1) + 2i) \;=\; (x - 1)^{2} - (2i)^{2}. $$ Compute: $(x - 1)^{2} = x^{2} - 2x + 1$ and $(2i)^{2} = 4i^{2} = -4$. Therefore计算：$(x - 1)^{2} = x^{2} - 2x + 1$，$(2i)^{2} = 4i^{2} = -4$。因此 $$ (x - 1)^{2} - (2i)^{2} \;=\; (x^{2} - 2x + 1) - (-4) \;=\; x^{2} - 2x + 5. $$ The coefficients are real (in fact integer) as the theorem promises. $\checkmark$如定理所言，系数都是实数（实际上是整数）。$\checkmark$

(c) Multiply by the real linear factor and expand乘上实一次因式并展开 M1·A1·A1

Since the third root is $x = 3$, the cubic with lead $1$ is因第三个根为 $x = 3$，首项为 $1$ 的三次式为 $$ p(x) \;=\; (x - 3)(x^{2} - 2x + 5). $$ Expand step by step:逐步展开： $$ \begin{aligned} p(x) &= x \cdot (x^{2} - 2x + 5) - 3 \cdot (x^{2} - 2x + 5) \\ &= (x^{3} - 2x^{2} + 5x) - (3x^{2} - 6x + 15) \\ &= x^{3} - 2x^{2} + 5x - 3x^{2} + 6x - 15 \\ &= x^{3} - 5x^{2} + 11x - 15. \end{aligned} $$ So $a = -5$, $b = 11$, $c = -15$, all integers as required.故 $a = -5$、$b = 11$、$c = -15$，均为整数，符合要求。
Sanity-check $p(3) = 0$:验算 $p(3) = 0$： $27 - 5(9) + 11(3) - 15 = 27 - 45 + 33 - 15 = 0$. $\checkmark$

(d) Geometric interpretation几何意义 A1

The graph of $y = p(x)$ has exactly one real $x$-intercept, at $x = 3$. The irreducible quadratic factor $x^{2} - 2x + 5$ has discriminant $(-2)^{2} - 4(5) = -16 < 0$, so it has no real zeros , its parabola lies entirely above the $x$-axis (lead $+1 > 0$, no crossings), which means $p$ does not vanish anywhere else on the real line.$y = p(x)$ 的图象只有一个实 $x$ 轴截距，即 $x = 3$。不可约二次因式 $x^{2} - 2x + 5$ 的判别式 $(-2)^{2} - 4(5) = -16 < 0$，故无实零点——其抛物线整体位于 $x$ 轴上方（首项 $+1 > 0$，无交点），意味着 $p$ 在实数轴上其它任何位置都不为零。
Real coefficients $\Rightarrow$ non-real roots in conjugate pairs $\Rightarrow$ odd-degree polynomials have at least one real root.实系数 $\Rightarrow$ 非实根成共轭对 $\Rightarrow$ 奇次多项式至少有一个实根。 The complex conjugate root theorem has three downstream consequences worth locking in: (1) a real-coefficient polynomial of odd degree must have an odd number of real roots (because non-real roots come in pairs), so cubics always have at least one real $x$-intercept , here, the $x = 3$. (2) Every real-coefficient polynomial factors over $\mathbb{R}$ as a product of real linear factors and real irreducible quadratics (discriminant $< 0$) , never higher-degree irreducibles. (3) The conjugate-pair product $(x - (a + bi))(x - (a - bi)) = x^{2} - 2ax + (a^{2} + b^{2})$ gives a real quadratic with discriminant $-(2b)^{2} < 0$, automatically irreducible. The "$(A - B)(A + B)$" trick with $A = x - a$, $B = bi$ is the speed move , never multiply the brackets term by term.复共轭根定理有三条值得锁定的下游推论：(1) 奇次实系数多项式必有奇数个实根（因非实根成对出现），故三次式恒有至少一个实 $x$ 轴截距——本题即 $x = 3$。(2) 每个实系数多项式在 $\mathbb{R}$ 上分解为实一次因式与实不可约二次因式（判别式 $< 0$）的乘积，绝不会有更高次的不可约因式。(3) 共轭对乘积 $(x - (a + bi))(x - (a - bi)) = x^{2} - 2ax + (a^{2} + b^{2})$ 给出判别式 $-(2b)^{2} < 0$ 的实二次式，自动不可约。$A = x - a$、$B = bi$ 的 $(A - B)(A + B)$ 技巧是提速利器——千万不要逐项相乘展开。
PART III  ·  MODELING / APPLIED · SOLUTIONS第三部分  ·  建模 / 应用 · 详解Universal · 32 marks通用 · 共 32 分

Section C · Modeling and ApplicationsC 部分 · 建模与应用

Q10MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §4 Polynomial Modeling多项式建模 · HSA-CED.A.1 [10 marks][10 分]

Open-top box: $V(x) = x(20 - 2x)(16 - 2x)$, $x$ in cm. (a) Expand + degree. (b) Practical domain. (c) Solve $V(x) = 384$. (d) Box dimensions for one solution.无盖纸盒：$V(x) = x(20 - 2x)(16 - 2x)$，$x$ 单位 cm。(a) 展开并写出次数。(b) 实际定义域。(c) 求解 $V(x) = 384$。(d) 任选一个解给出纸盒尺寸。

Answer:答案：  (a) $V(x) = 4x^{3} - 72x^{2} + 320x$, degree $3$次数 $3$  ·  (b) $x \in (0, 8)$  ·  (c) $x = 2$ and $x = 4$ (the third algebraic root $x = 12$ is outside the domain)$x = 2$ 与 $x = 4$（第三个代数根 $x = 12$ 在定义域外）  ·  (d) e.g. $x = 2$: $L \times W \times H = 16 \times 12 \times 2$ cm例如 $x = 2$：长 $\times$ 宽 $\times$ 高 $= 16 \times 12 \times 2$ cm

(a) Expand $V(x)$ into standard form把 $V(x)$ 展开为标准形式 M1·A1

Multiply the two binomials first: $(20 - 2x)(16 - 2x) = 320 - 40x - 32x + 4x^{2} = 4x^{2} - 72x + 320$. Then先乘两个二项式：$(20 - 2x)(16 - 2x) = 320 - 40x - 32x + 4x^{2} = 4x^{2} - 72x + 320$。然后 $$ V(x) \;=\; x(4x^{2} - 72x + 320) \;=\; 4x^{3} - 72x^{2} + 320x. $$ Degree $= 3$ (cubic).次数 $= 3$（三次）。

(b) Practical domain实际定义域 A1·A1

Physical constraints: $x > 0$ (must cut a square of positive side), $20 - 2x > 0 \Rightarrow x < 10$ (long side stays positive), and $16 - 2x > 0 \Rightarrow x < 8$ (short side stays positive). The binding constraint is the smaller upper bound, $x < 8$. So $x \in (0, 8)$. Endpoints excluded: at $x = 0$ no box is cut (no height); at $x = 8$ the short side collapses to width $0$ , degenerate box.物理限制：$x > 0$（剪掉的正方形边长须为正）、$20 - 2x > 0 \Rightarrow x < 10$（长边须保持正值）、$16 - 2x > 0 \Rightarrow x < 8$（短边须保持正值）。生效约束是较小的上界 $x < 8$。故 $x \in (0, 8)$。端点排除原因：$x = 0$ 时不切（无高度）；$x = 8$ 时短边宽度坍缩为 $0$——退化纸盒。

(c) Solve $V(x) = 384$求解 $V(x) = 384$ M1·M1·A1·A1

Set $4x^{3} - 72x^{2} + 320x = 384$ and divide through by $4$:设 $4x^{3} - 72x^{2} + 320x = 384$，两边同除以 $4$： $$ x^{3} - 18x^{2} + 80x - 96 \;=\; 0. $$ Rational root theorem.有理根定理。 Lead $1$, constant $-96$; candidates are divisors of $96$ with both signs: $\pm \{1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96\}$.首项 $1$，常数项 $-96$；候选根为 $96$ 的正负因数：$\pm \{1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96\}$。
Test $x = 2$: $8 - 72 + 160 - 96 = 0$. $\checkmark$试 $x = 2$：$8 - 72 + 160 - 96 = 0$。$\checkmark$
Synthetic-divide by $(x - 2)$ on $1, -18, 80, -96$ with $c = 2$: bring down $1$; $1 \cdot 2 = 2$, $-18 + 2 = -16$; $-16 \cdot 2 = -32$, $80 - 32 = 48$; $48 \cdot 2 = 96$, $-96 + 96 = 0$.对 $1, -18, 80, -96$ 用 $(x - 2)$（$c = 2$）做综合除法：把 $1$ 拉下；$1 \cdot 2 = 2$，$-18 + 2 = -16$；$-16 \cdot 2 = -32$，$80 - 32 = 48$；$48 \cdot 2 = 96$，$-96 + 96 = 0$。
Depressed quadratic: $x^{2} - 16x + 48 = 0$. Solve by quadratic formula:降次二次式：$x^{2} - 16x + 48 = 0$。用求根公式： $$ x \;=\; \frac{16 \pm \sqrt{256 - 192}}{2} \;=\; \frac{16 \pm \sqrt{64}}{2} \;=\; \frac{16 \pm 8}{2} \;=\; 12 \;\text{or}\; 4. $$ Three algebraic roots: $x = 2, 4, 12$. Keep only those in $(0, 8)$: $x = 2$ and $x = 4$. (Reject $x = 12$.) Both are rational, so no rounding is needed.三个代数根：$x = 2, 4, 12$。仅保留 $(0, 8)$ 内的解：$x = 2$ 与 $x = 4$。（舍弃 $x = 12$。）两者皆为有理数，无需四舍五入。

(d) Box dimensions for one solution给出一个解对应的纸盒尺寸 A1·A1

Take $x = 2$. Length $= 20 - 2(2) = 16$ cm; width $= 16 - 2(2) = 12$ cm; height $= x = 2$ cm. So the box is $16 \times 12 \times 2$ cm. Volume check: $16 \cdot 12 \cdot 2 = 384$ cm$^{3}$. $\checkmark$ (The other valid solution $x = 4$ gives a $12 \times 8 \times 4$ box with the same volume.)取 $x = 2$。长 $= 20 - 2(2) = 16$ cm；宽 $= 16 - 2(2) = 12$ cm；高 $= x = 2$ cm。故纸盒尺寸为 $16 \times 12 \times 2$ cm。体积验算：$16 \cdot 12 \cdot 2 = 384$ cm$^{3}$。$\checkmark$（另一可行解 $x = 4$ 给出 $12 \times 8 \times 4$ 的盒子，体积相同。）
Always factor before expanding , factored form gives the practical domain for free.先因式后展开——因式形式直接给出实际定义域。 The temptation is to expand $V(x) = x(20 - 2x)(16 - 2x)$ first, then hunt for the domain. Backwards: the factored form is exactly the form where each binding inequality is one factor > 0, so the domain $\cap$ of the three positivity conditions falls out in two lines. The "rational root, then synthetic-divide, then quadratic formula" workflow is the universal cubic-equation pipeline for HSA-CED.A.1: it requires no advanced tools and is what AP graders mark step by step. The third root $x = 12$ is mathematically valid (substitute and check: $12^{3} - 18(144) + 80(12) - 96 = 1728 - 2592 + 960 - 96 = 0$ $\checkmark$) but lies outside the practical domain , a box whose corner cut exceeds half the short side is geometrically impossible. State the rejection reason explicitly to bank the A1.直觉上会想先把 $V(x) = x(20 - 2x)(16 - 2x)$ 展开再求定义域。顺序反了：因式形式恰好让每个约束条件对应"某因式 > 0"，三个正性条件的定义域交集两行就能写出来。"有理根定理 + 综合除法 + 求根公式"是 HSA-CED.A.1 三次方程的通用流水线，无需高级工具，正是 AP 阅卷逐步给分的姿势。第三个根 $x = 12$ 在数学上完全合法（代入验证：$12^{3} - 18(144) + 80(12) - 96 = 1728 - 2592 + 960 - 96 = 0$ $\checkmark$），但不在实际定义域内——切角边长超过短边一半几何上不可能。显式说明舍弃原因可以稳拿这一 A1。
Q11MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §5 Building a Polynomial from its Zeros由零点构造多项式 · MHF4U Polynomial & Rational FunctionsMHF4U 多项式与有理函数 [11 marks][11 分]

Cubic with simple zero $x = -2$, double zero $x = 1$, and $f(0) = 8$. (a) Factored form. (b) Find $a$. (c) Standard form. (d) End behaviour, multiplicity behaviour, sketch.三次式：单零点 $x = -2$、二重零点 $x = 1$、$f(0) = 8$。(a) 因式形式。(b) 求 $a$。(c) 标准形式。(d) 末端行为、重数行为、作图。

Answer:答案：  (a) $f(x) = a(x + 2)(x - 1)^{2}$  ·  (b) $a = 4$  ·  (c) $f(x) = 4x^{3} - 12x + 8$  ·  (d) ends $\to \pm \infty$ matching lead $+$; cross at $-2$, touch+bounce at $1$; sketch below两端 $\to \pm \infty$ 与正首项一致；$-2$ 处穿过，$1$ 处相切反弹；作图见下

(a) Factored form from zeros and multiplicities由零点与重数写出因式形式 M1·A1

A simple zero at $r = -2$ contributes the factor $(x - (-2)) = (x + 2)$ with multiplicity $1$; the double zero at $r = 1$ contributes $(x - 1)^{2}$. With unknown leading coefficient $a$:$r = -2$ 的单零点提供因式 $(x - (-2)) = (x + 2)$（重数 $1$）；$r = 1$ 的二重零点提供 $(x - 1)^{2}$。设首项系数为待定的 $a$： $$ f(x) \;=\; a(x + 2)(x - 1)^{2}. $$ Total degree $= 1 + 2 = 3$, matching "cubic" as stated. $\checkmark$总次数 $= 1 + 2 = 3$，与题述"三次"一致。$\checkmark$

(b) Determine $a$ from $f(0) = 8$由 $f(0) = 8$ 确定 $a$ M1·A1

$$ f(0) \;=\; a(0 + 2)(0 - 1)^{2} \;=\; a \cdot 2 \cdot 1 \;=\; 2a. $$ Set $2a = 8 \Rightarrow a = 4$.令 $2a = 8 \Rightarrow a = 4$。

(c) Expand to standard form展开为标准形式 M1·A1·A1

$(x - 1)^{2} = x^{2} - 2x + 1$, then $(x + 2)(x^{2} - 2x + 1) = x^{3} - 2x^{2} + x + 2x^{2} - 4x + 2 = x^{3} - 3x + 2$. Multiply by $a = 4$:$(x - 1)^{2} = x^{2} - 2x + 1$，再 $(x + 2)(x^{2} - 2x + 1) = x^{3} - 2x^{2} + x + 2x^{2} - 4x + 2 = x^{3} - 3x + 2$。乘以 $a = 4$： $$ f(x) \;=\; 4(x^{3} - 3x + 2) \;=\; 4x^{3} - 12x + 8. $$ Note the $x^{2}$-coefficient is $0$ , the linear and cubic terms expand cleanly with no quadratic term. Audit: $f(0) = 8$ $\checkmark$; $f(1) = 4 - 12 + 8 = 0$ $\checkmark$; $f(-2) = -32 + 24 + 8 = 0$ $\checkmark$.注意 $x^{2}$ 系数为 $0$——一次项与三次项展开后整齐相消，正好没有二次项。审核：$f(0) = 8$ $\checkmark$；$f(1) = 4 - 12 + 8 = 0$ $\checkmark$；$f(-2) = -32 + 24 + 8 = 0$ $\checkmark$。

(d) End behaviour, multiplicity behaviour, sketch末端行为、重数行为、作图 A1·A1·A1·A1

End behaviour.末端行为。 Degree $3$ (odd), lead $+4 > 0$. So $f(x) \to -\infty$ as $x \to -\infty$ and $f(x) \to +\infty$ as $x \to +\infty$ (down-on-left, up-on-right).次数 $3$（奇），首项 $+4 > 0$。故 $x \to -\infty$ 时 $f(x) \to -\infty$，$x \to +\infty$ 时 $f(x) \to +\infty$（左下右上）。
Multiplicity behaviour at each zero.各零点处的重数行为。 At $x = -2$ (mult $1$, odd): graph crosses the $x$-axis. At $x = 1$ (mult $2$, even): graph touches the axis and bounces back (no sign change).在 $x = -2$（重数 $1$，奇）：图象穿过 $x$ 轴。在 $x = 1$（重数 $2$，偶）：图象相切于 $x$ 轴并反弹（不变号）。
Sketch.作图。 Coming from $-\infty$ on the far left, $f$ rises, crosses the $x$-axis at $(-2, 0)$ into the upper half-plane, peaks somewhere in $(-2, 1)$ (passing through the $y$-intercept $(0, 8)$ , in fact this is near the local max), descends, touches the $x$-axis at $(1, 0)$ without crossing, and continues climbing to $+\infty$ on the right. The graph stays $\ge 0$ on $[-2, \infty)$ , the bounce at $x = 1$ keeps $f$ on the same (positive) side.最左侧自 $-\infty$ 升起，$f$ 在 $(-2, 0)$ 处穿过 $x$ 轴进入上半平面，在 $(-2, 1)$ 间某处达到峰值（经过 $y$ 轴截距 $(0, 8)$——其实正在局部极大附近），下降，在 $(1, 0)$ 处与 $x$ 轴相切但不穿过，再继续上升至 $+\infty$。图象在 $[-2, \infty)$ 上保持 $\ge 0$——$x = 1$ 处的反弹使 $f$ 始终位于同一（正）侧。
Build factored form from zeros first, then pin down $a$ with the remaining condition.先由零点写因式形式，再用剩余条件锁定 $a$。 Reverse-engineering a polynomial from its graph data is a two-step recipe: (1) every zero $r$ with multiplicity $m$ contributes $(x - r)^{m}$; multiply them all and prefix an unknown leading coefficient $a$; (2) use the one "extra" datum (here $f(0) = 8$) to solve for $a$. The order matters: if you try to find $a$ before writing all the factors, you have no equation to solve. Sign trap: signed zero $x = -2$ becomes factor $(x + 2)$, not $(x - 2)$ , the bracket sign flips relative to the root. MHF4U markers reward students who explicitly link odd/even multiplicity to cross/touch and end behaviour; writing "the graph crosses at $-2$ because the factor $(x + 2)$ has multiplicity $1$" earns an A1 that "the graph touches at $1$" alone does not.由图象信息逆向构造多项式是两步流程：(1) 每个重数为 $m$ 的零点 $r$ 贡献 $(x - r)^{m}$；全部相乘再加未知首项系数 $a$；(2) 用"额外"的一个条件（本题 $f(0) = 8$）解出 $a$。顺序很关键：若在写齐因式前就找 $a$，根本没有方程可解。符号陷阱：带符号的零点 $x = -2$ 对应因式 $(x + 2)$，而不是 $(x - 2)$——括号内符号与根相反。MHF4U 阅卷会奖励显式将奇 / 偶重数与穿过 / 相切及末端行为挂钩的学生；写"图象在 $-2$ 处穿过，因为因式 $(x + 2)$ 重数为 $1$"能拿到一个 A1，只写"图象在 $1$ 处相切"就拿不到。
Q12HARD难Honors荣誉级 🇨🇦 BC卑 🇺🇸 US美 BC Provincial-style卑诗省考风格 §7 Polynomial Inequality (Sign Analysis)多项式不等式（符号分析） · BC PC12 / HSA-REI.D.11 [11 marks][11 分]

$x^{3} - 4x^{2} + x + 6 \le 0$. (a) Verify $x = -1$ is a root. (b) Full factor + zeros. (c) Sign chart. (d) Solution set. (e) Reservoir context: $r(t) = -p(t) \ge 0$ on $[0, 4]$.$x^{3} - 4x^{2} + x + 6 \le 0$。(a) 验证 $x = -1$ 是根。(b) 完全因式分解 + 零点。(c) 符号表。(d) 解集。(e) 水库情境：$r(t) = -p(t) \ge 0$ 在 $[0, 4]$ 上。

Answer:答案：  (a) $p(-1) = 0 \Rightarrow (x + 1)$ is a factor$p(-1) = 0 \Rightarrow (x + 1)$ 是因式  ·  (b) $p(x) = (x + 1)(x - 2)(x - 3)$; zeros零点 $-1, 2, 3$  ·  (c) $-, +, -, +$ on $(-\infty, -1), (-1, 2), (2, 3), (3, \infty)$在 $(-\infty, -1), (-1, 2), (2, 3), (3, \infty)$ 上依次为 $-, +, -, +$  ·  (d) $(-\infty, -1] \cup [2, 3]$  ·  (e) $t \in [2, 3]$ (hours after midnight)$t \in [2, 3]$（零点后的小时数）

(a) Verify $x = -1$ is a root验证 $x = -1$ 是根 M1·A1

$$ p(-1) \;=\; (-1)^{3} - 4(-1)^{2} + (-1) + 6 \;=\; -1 - 4 - 1 + 6 \;=\; 0. $$ Since $p(-1) = 0$, by the factor theorem $(x - (-1)) = (x + 1)$ is a factor of $p(x)$.由于 $p(-1) = 0$，根据因式定理，$(x - (-1)) = (x + 1)$ 是 $p(x)$ 的因式。

(b) Full factorisation and zeros完全因式分解与零点 M1·A1·A1

Synthetic-divide $1, -4, 1, 6$ by $c = -1$: bring down $1$; $1 \cdot (-1) = -1$, $-4 + (-1) = -5$; $-5 \cdot (-1) = 5$, $1 + 5 = 6$; $6 \cdot (-1) = -6$, $6 + (-6) = 0$.对 $1, -4, 1, 6$ 用 $c = -1$ 做综合除法：把 $1$ 拉下；$1 \cdot (-1) = -1$，$-4 + (-1) = -5$；$-5 \cdot (-1) = 5$，$1 + 5 = 6$；$6 \cdot (-1) = -6$，$6 + (-6) = 0$。
Depressed quadratic: $x^{2} - 5x + 6 = (x - 2)(x - 3)$. So降次二次式：$x^{2} - 5x + 6 = (x - 2)(x - 3)$。故 $$ p(x) \;=\; (x + 1)(x - 2)(x - 3). $$ Three real zeros: $x = -1$, $x = 2$, $x = 3$.三个实零点：$x = -1$、$x = 2$、$x = 3$。

(c) Sign chart符号表 M1·A1·A1

The three zeros partition the real line into four intervals. Test one point per interval in $p(x) = (x + 1)(x - 2)(x - 3)$:三个零点把实轴分成四段。在 $p(x) = (x + 1)(x - 2)(x - 3)$ 中每段选一点测试：
  • $x = -2$: $(-2 + 1)(-2 - 2)(-2 - 3) = (-1)(-4)(-5) = -20$. Sign:符号： $-$.
  • $x = 0$: $(1)(-2)(-3) = 6$. Sign:符号： $+$.
  • $x = 2.5$: $(3.5)(0.5)(-0.5) = -0.875$. Sign:符号： $-$.
  • $x = 4$: $(5)(2)(1) = 10$. Sign:符号： $+$.
Sign chart (left to right): $\;-\;\big|\;-1\;\big|\;+\;\big|\;2\;\big|\;-\;\big|\;3\;\big|\;+\;$. Each odd-multiplicity zero flips the sign; the alternating $-, +, -, +$ pattern is exactly what symmetry predicts for a cubic with positive lead and three simple real zeros (the graph rises from $-\infty$, crosses up at $-1$, crosses down at $2$, crosses up at $3$, climbs to $+\infty$).符号表（自左向右）：$\;-\;\big|\;-1\;\big|\;+\;\big|\;2\;\big|\;-\;\big|\;3\;\big|\;+\;$。每个奇重数零点处符号翻转；正首项 + 三个单实零点的三次式必然呈 $-, +, -, +$ 交替（图象从 $-\infty$ 升起，在 $-1$ 上穿，在 $2$ 下穿，在 $3$ 上穿，升至 $+\infty$）——这是对称性给出的强预测。

(d) Solution set of $p(x) \le 0$$p(x) \le 0$ 的解集 A1·A1

Non-strict $\le 0$: take the $-$ intervals and the boundary zeros (where $p = 0$). From (c) the $-$ intervals are $(-\infty, -1)$ and $(2, 3)$; adjoin the zeros $-1, 2, 3$:非严格不等式 $\le 0$：取所有 $-$ 段并边界零点（$p = 0$ 处）。由 (c)，$-$ 段为 $(-\infty, -1)$ 与 $(2, 3)$；并入零点 $-1, 2, 3$： $$ \{x : p(x) \le 0\} \;=\; (-\infty, -1] \cup [2, 3]. $$

(e) Reservoir context水库情境 A1

$r(t) = -p(t)$, so $r(t) \ge 0 \Leftrightarrow p(t) \le 0$. On the situational domain $t \in [0, 4]$, intersect with the solution from (d): $\bigl((-\infty, -1] \cup [2, 3]\bigr) \cap [0, 4] = [2, 3]$. Water flows into the reservoir on the subinterval $\boxed{t \in [2, 3]}$ , from 2 a.m. to 3 a.m. (the rest of the four-hour window the net flow is out of the reservoir).$r(t) = -p(t)$，因此 $r(t) \ge 0 \Leftrightarrow p(t) \le 0$。在情境定义域 $t \in [0, 4]$ 上与 (d) 的解集求交：$\bigl((-\infty, -1] \cup [2, 3]\bigr) \cap [0, 4] = [2, 3]$。水流入水库的子区间为 $\boxed{t \in [2, 3]}$——即凌晨 2 点到 3 点（其余 4 小时窗口内净流向是流出）。
Sign chart $=$ one diagram, all four inequalities for free.一张符号表 $=$ 四种不等式全部免费。 Once $p$ is fully factored and the sign chart is drawn, every related inequality reads off without re-work: $p < 0$ takes the open $-$ intervals; $p \le 0$ takes the closed $-$ intervals (including the zeros); $p > 0$ takes the open $+$ intervals; $p \ge 0$ takes the closed $+$ intervals. The $\le$ vs $<$ distinction only changes whether the zeros are bracketed. Two structural moves to lock in: (1) cubic with positive lead and three distinct real zeros always alternates $-, +, -, +$ , you can read the sign chart off the leading-coefficient sign alone, then audit with one test point; (2) negating $p$ flips the sign chart wholesale ($+ \leftrightarrow -$), so part (e)'s "$r = -p \ge 0$" is identical to "$p \le 0$" , no second sign chart needed. BC PC12 markers reward this kind of structural reuse with an R1; a re-drawn sign chart for $r$ wastes time. Caution at boundary: open vs closed brackets are an A1 trap , lose a mark by writing $(2, 3)$ instead of $[2, 3]$.一旦 $p$ 完全因式分解、符号表画好，所有相关不等式都能不再重算地直接读出：$p < 0$ 取开 $-$ 段；$p \le 0$ 取闭 $-$ 段（含零点）；$p > 0$ 取开 $+$ 段；$p \ge 0$ 取闭 $+$ 段。$\le$ 与 $<$ 的区别仅在于是否方括号包含零点。两条结构性招数务必锁定：(1) 正首项 + 三个不同实零点的三次式必呈 $-, +, -, +$ 交替——单看首项符号即可写出符号表，再用一个试点审核；(2) 对 $p$ 取负把符号表整体翻转（$+ \leftrightarrow -$），所以 (e) 的"$r = -p \ge 0$"等价于"$p \le 0$"——无需另作一张符号表。卑诗 PC12 阅卷会以 R1 奖励这种结构复用；再画一张 $r$ 的符号表是浪费时间。边界小心：开括号 / 闭括号是经典的 A1 陷阱——写成 $(2, 3)$ 而非 $[2, 3]$ 会丢分。