Algebra (modulus and polynomials)
مُقدَّم| English | المعنى |
|---|---|
| modulus/ˈmɒdjʊləs/ | القيمة المطلقة |
| remainder theorem/rɪˈmeɪndə ˈθɪərəm/ | نظرية الباقي |
| polynomial/ˌpɒlɪˈnəʊmɪəl/ | كثير الحدود |
| degree/dɪˈɡriː/ | الدرجة |
| factor theorem/ˈfæktə ˈθɪərəm/ | نظرية العامل |
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.
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 $x = -2$.
The modulus function
y = a|x − b| + c
The modulus makes a V-shape. Move its القمة (الرأس) with b and c; change how steep the arms are with a.
Solve |x − 2| < 3. The solution is −1 < x < ? What is the upper bound?
الصعوبة 2/5 · 140 عملة + مكافأة صغيرة عند الإكمال الأول اليوم
|x − 2| < 3 means 2 − 3 < x < 2 + 3, i.e. −1 < x < 5.
Solve |2x − 3| = 7. What is the larger solution?
الصعوبة 2/5 · 140 عملة + مكافأة صغيرة عند الإكمال الأول اليوم
2x − 3 = 7 → x = 5; or 2x − 3 = −7 → x = −2. Larger is 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
By the remainder theorem, what is the remainder when p(x) = x² + 3x + 2 is divided by (x − 1)?
الصعوبة 2/5 · 140 عملة + مكافأة صغيرة عند الإكمال الأول اليوم
Remainder = p(1) = 1 + 3 + 2 = 6.
(x − a) is a factor of p(x) exactly when:
الصعوبة 1/5 · 100 عملة + مكافأة صغيرة عند الإكمال الأول اليوم
The factor theorem: (x − a) is a factor if and only if 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
If p(3) = 0, then (x − 3) is a factor of p(x).
الصعوبة 1/5 · 100 عملة + مكافأة صغيرة عند الإكمال الأول اليوم
By the factor theorem, p(a) = 0 means (x − a) is a factor.
Given p(x) = x³ − 6x² + 11x − 6 and p(1) = 0, factorise: p(x) = (x − 1)(x² + bx + c). What is c?
الصعوبة 2/5 · 140 عملة + مكافأة صغيرة عند الإكمال الأول اليوم
Dividing x³ − 6x² + 11x − 6 by (x − 1) gives x² − 5x + 6, so 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.
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
Write from the solutions 3 and -7
- Solve the modulus equation absolute value of x + 2 equals 5.
Solve the equation given by the absolute value of x + 2 equals 5.
3 plus 2 is 5, and -7 plus 2 is -5. Both have absolute value 5.
What are the solutions?
الصعوبة 1/5 · 100 عملة + مكافأة صغيرة عند الإكمال الأول اليوم
The two branches are x = 3 and x = -7.