Algebra (modulus and polynomials) · 代数(模与多项式)
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| modulus/ˈmɒdjʊləs/ | 绝对值 | jué duì zhí |
| remainder theorem/rɪˈmeɪndə ˈθɪərəm/ | 余数定理 | yú shù dìng lǐ |
| polynomial/ˌpɒlɪˈnəʊmɪəl/ | 多项式 | duō xiàng shì |
| degree/dɪˈɡriː/ | 次数 | cì shù |
| factor theorem/ˈfæktə ˈθɪərəm/ | 因式定理 | yīn shì dìng lǐ |
The absolute value that saves bridges
- Engineers designing suspension cables need to know the maximum deviation from the ideal — whether above or below.
- The modulus 绝对值 (absolute value) measures distance from zero, ignoring direction. It appears everywhere: error bounds, tolerances, and inequalities.
拯救桥梁的绝对值
- 设计悬索的工程师需要知道与理想值的最大偏差——无论是高于还是低于。
- 模(modulus,绝对值)测量到零的距离,忽略方向。它到处出现:误差界限、公差和不等式。
The modulus function
- The modulus $|x|$ is the size of a number, sign removed ($|x| \geq 0$). Its graph is a "V".
- Useful rules:
- $|a| = |b| \Leftrightarrow a^2 = b^2$,
- $|x - a| < b \Leftrightarrow a - b < x < a + b$.
The modulus function $y = |x - 2|$ creates a V-shape: everything below the axis is reflected above it.
Worked example. Solve $|2x - 3| = 7$. Either $2x - 3 = 7 \Rightarrow x = 5$, or $2x - 3 = -7 \Rightarrow x = -2$. Solutions: $x = 5$ or $x = -2$.
模函数
- 模 $|x|$ 是一个数的大小,去掉符号($|x| \geq 0$)。它的图是一个“V”。
- 有用的规则:
- $|a| = |b| \Leftrightarrow a^2 = b^2$,
- $|x - a| < b \Leftrightarrow a - b < x < a + b$。
模函数 $y = |x - 2|$ 形成一个 V 形:轴下面的一切都被反射到上面。
算例。 解 $|2x - 3| = 7$。要么 $2x - 3 = 7 \Rightarrow x = 5$,要么 $2x - 3 = -7 \Rightarrow x = -2$。解:$x = 5$ 或 $x = -2$。
The modulus function · 模函数
y = a|x − b| + c
The modulus makes a V-shape. Move its vertex · 顶点 with b and c; change how steep the arms are with a. · 模做出一个 V 形。用 b 和 c 移动它的顶点;用 a 改变两臂的陡度。
Solve |x − 2| < 3. The solution is −1 < x < ? What is the upper bound? · 解 |x − 2| < 3。解是 −1 < x < ?上界是多少?
|x − 2| < 3 means 2 − 3 < x < 2 + 3, i.e. −1 < x < 5. · |x − 2| < 3 意味着 2 − 3 < x < 2 + 3,即 −1 < x < 5。
Solve |2x − 3| = 7. What is the larger solution? · 解 |2x − 3| = 7。较大的解是多少?
2x − 3 = 7 → x = 5; or 2x − 3 = −7 → x = −2. Larger is 5. · 2x − 3 = 7 → x = 5;或 2x − 3 = −7 → x = −2。较大的是 5。
Factor & remainder theorems 余数定理
- A polynomial 多项式 is a sum of powers of $x$; its degree 次数 is the highest power.
- Remainder theorem: the remainder when $p(x) \div (x - a)$ is $p(a)$.
- Factor theorem 因式定理: $(x - a)$ is a factor exactly when $p(a) = 0$.
Don't confuse the two theorems. The remainder theorem gives the remainder for any divisor $(x - a)$. The factor theorem is a special case: when the remainder is zero, $(x - a)$ is a factor.
Taking the modulus folds the negative part of the line up into a V
因式定理与余数定理
- 一个多项式(polynomial)是 $x$ 的幂的和;它的次数(degree)是最高的幂。
- 余数定理(remainder theorem):$p(x) \div (x - a)$ 的余数是 $p(a)$。
- 因式定理(factor theorem):$(x - a)$ 是一个因式恰好当 $p(a) = 0$。
不要混淆这两个定理。 余数定理给出对任何除式 $(x - a)$ 的余数。因式定理是一个特殊情况:当余数是零时,$(x - a)$ 是一个因式。
取模把直线的负部分向上折成一个 V
By the remainder theorem, what is the remainder when p(x) = x² + 3x + 2 is divided by (x − 1)? · 由余数定理,p(x) = x² + 3x + 2 除以 (x − 1) 的余数是多少?
Remainder = p(1) = 1 + 3 + 2 = 6. · 余数 = p(1) = 1 + 3 + 2 = 6。
(x − a) is a factor of p(x) exactly when: · (x − a) 是 p(x) 的一个因式,恰好当:
The factor theorem: (x − a) is a factor if and only if p(a) = 0. · 因式定理:(x − a) 是一个因式当且仅当 p(a) = 0。
Worked example — factor theorem
- Show that $(x - 2)$ is a factor of $p(x) = x^3 - 5x^2 + 2x + 8$.
- $p(2) = 8 - 20 + 4 + 8 = 0$. Since $p(2) = 0$, $(x - 2)$ is a factor.
- Divide: $x^3 - 5x^2 + 2x + 8 = (x - 2)(x^2 - 3x - 4) = (x - 2)(x - 4)(x + 1)$.
Opposite signs of f(a) and f(b) trap a root between a and b
算例——因式定理
- 证明 $(x - 2)$ 是 $p(x) = x^3 - 5x^2 + 2x + 8$ 的一个因式。
- $p(2) = 8 - 20 + 4 + 8 = 0$。因为 $p(2) = 0$,$(x - 2)$ 是一个因式。
- 相除:$x^3 - 5x^2 + 2x + 8 = (x - 2)(x^2 - 3x - 4) = (x - 2)(x - 4)(x + 1)$。
f(a) 和 f(b) 的相反符号把一个根困在 a 和 b 之间
If p(3) = 0, then (x − 3) is a factor of p(x). · 如果 p(3) = 0,那么 (x − 3) 是 p(x) 的一个因式。
By the factor theorem, p(a) = 0 means (x − a) is a factor. · 由因式定理,p(a) = 0 意味着 (x − a) 是一个因式。
Given p(x) = x³ − 6x² + 11x − 6 and p(1) = 0, factorise: p(x) = (x − 1)(x² + bx + c). What is c? · 给定 p(x) = x³ − 6x² + 11x − 6 且 p(1) = 0,因式分解:p(x) = (x − 1)(x² + bx + c)。c 是多少?
Dividing x³ − 6x² + 11x − 6 by (x − 1) gives x² − 5x + 6, so c = 6. · x³ − 6x² + 11x − 6 除以 (x − 1) 给出 x² − 5x + 6,所以 c = 6。
Solving polynomial equations
- Use the factor theorem to find one root, then divide to reduce the degree.
- A cubic has at most 3 real roots; a quartic at most 4.
- Sketch the graph to check: the number of $x$-axis crossings matches the number of real roots.
- Differentiate a fraction with the quotient rule.
求解多项式方程
- 用因式定理找到一个根,然后相除以降低次数。
- 一个三次式最多有 3 个实根;一个四次式最多 4 个。
- 画图来检查:$x$ 轴交叉的数量匹配实根的数量。
- 用商(quotient)法则对分式求导。
You've got it
- $|x|$ removes the sign; $|x - a| < b \Leftrightarrow a - b < x < a + b$
- remainder when $p(x) \div (x-a)$ is $p(a)$
- factor theorem: $(x-a)$ is a factor $\Leftrightarrow p(a) = 0$
- find one root, divide, then solve the remaining quadratic
你掌握了
- $|x|$ 去掉符号;$|x - a| < b \Leftrightarrow a - b < x < a + b$
- $p(x) \div (x-a)$ 的余数是 $p(a)$
- 因式定理:$(x-a)$ 是一个因式 $\Leftrightarrow p(a) = 0$
- 找到一个根,相除,然后解剩下的二次式