📚 Polynomial Equations of Degree n | n 次多项式方程
A Level Mathematics often extends quadratic methods to polynomial equations of degree 3, 4 or higher. This AQA revision guide covers the standard form, root theorems, division, complex roots, Vieta's formulas, repeated roots and inequalities.
A Level 数学经常把二次方程的方法推广到三次、四次或更高次的多项式方程。这份 AQA 复习指南涵盖标准形式、根定理、除法、复数根、韦达公式、重根以及不等式。
1. Standard Form and Degree | 标准形式与次数
re real or complex numbers. The degree is the highest power of x with a non-zero coefficient.
n 次多项式方程的一般形式为 aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀ = 0，其中 aₙ ≠ 0，系数为实数或复数。次数就是指系数非零的 x 的最高次幂。
has degree 5, while (x - 1)²(x + 2) = 0 expands to a cubic equation with a repeated root.
例如，3x⁵ - 2x³ + x - 7 = 0 是五次方程，而 (x - 1)²(x + 2) = 0 展开后是一个带有重根的三次方程。
2. Fundamental Theorem of Algebra | 代数基本定理
The Fundamental Theorem of Algebra states that every non-constant polynomial equation with real or complex coefficients has at least one complex root. It follows that a degree n polynomial equation has exactly n complex roots, counting multiplicity.
代数基本定理指出：每个系数为实数或复数的非常数多项式方程至少有一个复数根。由此可知，一个 n 次多项式方程恰好有 n 个复数根（重根按重数计算）。
For example, x³ - 3x + 2 = 0 has roots 1, 1 and -2, so the root x = 1 has multiplicity 2.
这并不意味着所有根都互不相同；根可以重复。例如 x³ - 3x + 2 = 0 的根为 1、1 和 -2，因此根 x = 1 的重数为 2。
3. Factor Theorem and Remainder Theorem | 因式定理与余数定理
of f(x), and conversely if (x - a) is a factor then f(a) = 0. This is the Factor Theorem.
对于多项式 f(x)，如果 f(a) = 0，那么 (x - a) 就是 f(x) 的一个因式；反过来，如果 (x - a) 是因式，则 f(a) = 0。这就是因式定理。
The Remainder Theorem states that when f(x) is divided by (x - a), the remainder is f(a). This is useful for testing possible linear factors quickly.
余数定理指出：当 f(x) 除以 (x - a) 时，余数为 f(a)。它可以帮助我们快速检验可能的线性因式。
f(a) = 0 ⇔ (x - a) is a factor of f(x)
f(a) = 0 ⇔ (x - a) 是 f(x) 的因式
4. Polynomial Division | 多项式除法
Polynomial long division or synthetic division reduces the degree of a polynomial after one factor is known. Division is especially useful when solving cubics or quartics.
在已知一个因式后，可以使用多项式长除法或综合除法降低多项式次数。除法在求解三次或四次方程时特别有用。
Synthetic division with root 2 gives quotient x² - 2x - 3, so f(x) = (x - 2)(x² - 2x - 3).
示例：用 (x - 2) 除 f(x) = x³ - 4x² + x + 6。以根 2 进行综合除法得到商 x² - 2x - 3，因此 f(x) = (x - 2)(x² - 2x - 3)。
The quadratic then factorises as (x - 3)(x + 1), so the roots are 2, 3 and -1.
该二次式可分解为 (x - 3)(x + 1)，所以根为 2、3 和 -1。
5. Solving Cubic Equations | 求解三次方程
To solve a cubic equation, first use the Factor Theorem to find one rational root, then divide by the corresponding linear factor to obtain a quadratic. Solve the quadratic by factorising, completing the square or the quadratic formula.
要求解三次方程，先使用因式定理找到一个有理根，然后除以相应的线性因式得到一个二次式。再用因式分解、配方法或求根公式解该二次式。
is a factor. Division gives x² - 5x + 6 = (x - 2)(x - 3), hence the roots are 1, 2 and 3.
示例：解 x³ - 6x² + 11x - 6 = 0。检验 x = 1 得 1 - 6 + 11 - 6 = 0，所以 (x - 1) 是因式。除法得 x² - 5x + 6 = (x - 2)(x - 3)，因此根为 1、2 和 3。
If no rational root is apparent, AQA questions often give one root or an additional condition to make the cubic solvable.
如果没有明显的实有理根，AQA 试题通常会给出一个根或附加条件，使三次方程可以求解。
6. Solving Quartic Equations | 求解四次方程
of the form ax⁴ + bx² + c = 0, let y = x², solve ay² + by + c = 0, then take square roots.
某些四次方程可以通过代换降为二次方程。如果四次方程形如 ax⁴ + bx² + c = 0，令 y = x²，解 ay² + by + c = 0，再开平方即可。
5x² + 4 = 0. Let y = x², giving y² - 5y + 4 = 0, so y = 1 or y = 4. Thus x = ±1 or x = ±2.
示例：解 x⁴ - 5x² + 4 = 0。令 y = x²，得 y² - 5y + 4 = 0，所以 y = 1 或 y = 4。于是 x = ±1 或 x = ±2。
Other quartics may be solved by finding two quadratic factors, often after using the Factor Theorem twice.
其他四次方程可以通过找出两个二次因式来求解，通常需要连续使用两次因式定理。
7. Complex Roots and Conjugate Pairs | 复数根与共轭根对
If a polynomial has real coefficients, any non-real complex root must occur in a conjugate pair. That is, if a + bi is a root, then a - bi is also a root.
如果一个多项式的系数为实数，那么任何非实复数根都必须以共轭对的形式出现。也就是说，如果 a + bi 是根，那么 a - bi 也是根。
0 has a root 2 + i, the conjugate 2 - i must also be a root. The sum of these two roots is 4, so the remaining root is 2 because the total sum of roots is 6.
示例：已知 x³ - 6x² + 13x - 10 = 0 有一个根为 2 + i，则共轭根 2 - i 也必为根。这两个根的和为 4，由于三个根之和为 6，因此剩余根为 2。
ten arise after a quadratic factor has no real roots, for example x² + 1 = 0 gives x = ±i.
复数根通常出现在二次因式没有实根时，例如 x² + 1 = 0 给出 x = ±i。
8. Sum and Product of Roots | 根的和与积
Vieta' ial to symmetric sums of its roots. For a cubic ax³ + bx² + cx + d = 0 with roots α, β, γ:
韦达公式将多项式系数与根的对称和联系起来。对于三次方程 ax³ + bx² + cx + d = 0，其根为 α、β、γ，则：
α + β + γ = −b/a, αβ + αγ + βγ = c/a, αβγ = −d/a
α + β + γ = −b/a，αβ + αγ + βγ = c/a，αβγ = −d/a
For a quartic ax⁴ + bx³ + cx² + dx + e = 0 with roots α, β, γ, δ, the main formulas are:
对于四次方程 ax⁴ + bx³ + cx² + dx + e = 0，其根为 α、β、γ、δ，主要公式为：
| Sum of roots | α + β + γ + δ = −b/a |
| Sum of product pairs | αβ + αγ + αδ + βγ + βδ + γδ = c/a |
| Sum of product triples | αβγ + αβδ + αγδ + βγδ = −d/a |
| Product of roots | αβγδ = e/a |
These are very useful for finding unknown roots when one or more roots are given, or for forming a polynomial from its roots.
当已知部分根或需要由根构造多项式时，这些公式非常有用。
9. Repeated Roots and Graph Behaviour | 重根与图像行为
A polynomial root has multiplicity equal to the power of its corresponding factor. If (x - r)ᵏ is a factor, then r is a root of multiplicity k.
多项式根的重数等于其对应因式的幂次。如果 (x - r)ᵏ 是因式，那么 r 就是重数为 k 的根。
= r. For even multiplicity, the graph touches the x-axis and turns back without crossing.
重数为奇数时，图像在 x = r 处穿过 x 轴；重数为偶数时，图像在 x = r 处接触 x 轴并返回，不穿过 x 轴。
es the x-axis there, and a root at x = -2 with multiplicity 1, so the graph crosses there.
示例：f(x) = (x - 1)²(x + 2) 在 x = 1 处有重数为 2 的根，因此图像在此处接触 x 轴；在 x = -2 处有重数为 1 的根，因此图像在此处穿过 x 轴。
At a repeated root of multiplicity at least 2, the derivative f'(x) is also zero. This connects to stationary points on the graph.
在重数至少为 2 的重根处，导数 f'(x) 也为零。这与图像上的驻点有关。
10. Polynomial Inequalities | 多项式不等式
l roots of f(x) = 0, then create a sign table over the intervals separated by these roots.
要求解多项式不等式如 f(x) > 0，先求出 f(x) = 0 的所有实根，然后在由这些根分隔的区间上建立符号表。
Example: Solve x³ - 4x > 0. Factorise to x(x - 2)(x + 2) > 0. The critical values are -2, 0 and 2. Testing intervals shows the solution is -2 < x < 0 or x > 2.
示例：解 x³ - 4x > 0。分解得 x(x - 2)(x + 2) > 0。临界值为 -2、0 和 2。检验区间后可知解为 -2 < x < 0 或 x > 2。
At a repeated root of even multiplicity, the sign does not change; at a root of odd multiplicity, the sign changes.
在重数为偶数的重根处，符号不变；在重数为奇数的根处，符号改变。
11. Exam Tips and Common Mistakes | 考试技巧与常见错误
Always write a polynomial in descending powers before dividing or applying Vieta's formulas. Check that the leading coefficient is not zero, otherwise the degree is lower than expected.
在进行除法或使用韦达公式之前，务必按降幂书写多项式。检查首项系数是否为零，否则次数会低于预期。
Do not forget that a degree n equation has n roots counting multiplicity; missing a repeated root is a common error. When solving by substitution, remember to reverse the substitution to find x.
不要忘记 n 次方程有 n 个根（按重数计算）；遗漏重根是常见错误。使用代换法求解时，记得将代换反解回 x。
For complex roots, always include the conjugate pair when the coefficients are real. In inequalities, sketch the graph or use a sign table rather than using division by x, which can reverse inequality signs incorrectly.
对于复数根，当系数为实数时一定要列出共轭根。解不等式时，尽量画图或使用符号表，不要直接除以 x，因为那样可能错误地改变不等号方向。
Finally, verify your roots by substitution into the original equation, especially when a question asks for exact values or all solutions.
最后，把根代回原方程进行验证，尤其是题目要求精确值或所有解时。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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