Polynomial Equations of Degree n | n 次多项式方程

📚 Polynomial Equations of Degree n | n 次多项式方程

A polynomial equation of degree n is an equation that can be written in the form (P(x)=0), where (P(x)) is a polynomial whose highest power of (x) is (n). This equation is fundamental to much of A-Level mathematics, and understanding how to find its roots is essential for solving real-world problems and for success in examinations.

一个次数为 (n) 的多项式方程是形如 (P(x)=0) 的方程，其中 (P(x)) 是最高次项为 (x) 的 (n) 次方的多项式。这个方程是 A-Level 数学的基础内容，理解如何求它的根，对于解决实际问题以及在考试中取得好成绩都至关重要。

The general form of a polynomial equation of degree (n) is shown below. Here, (a_n neq 0), and the coefficients (a_n, a_{n-1}, dots, a_0) are real numbers. The number (n) must be a non-negative integer.

(n) 次多项式方程的一般形式如下所示。其中 (a_n neq 0)，系数 (a_n, a_{n-1}, dots, a_0) 是实数，次数 (n) 是非负整数。

aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀ = 0

According to the Fundamental Theorem of Algebra, a polynomial equation of degree (n) has exactly (n) roots when complex numbers are counted, provided the roots are counted with multiplicity. For AQA A-Level Mathematics, you will often work with real roots and simple complex roots, and you must be able to relate roots to factors.

根据代数基本定理，如果把复数根也计算在内，并且按重数计数，那么 (n) 次多项式方程恰好有 (n) 个根。在 AQA A-Level 数学中，你经常处理实根和简单的复数根，并且必须能够把根与因式联系起来。

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1. Roots, Factors and the Factor Theorem | 根、因式与因式定理

For any polynomial (P(x)), the real number (a) is a root of the equation (P(x)=0) if and only if ((x-a)) is a factor of (P(x)). This result is called the Factor Theorem, and it is one of the most powerful tools for solving polynomial equations.

对于任意多项式 (P(x))，实数 (a) 是方程 (P(x)=0) 的根，当且仅当 ((x-a)) 是 (P(x)) 的因式。这个结论称为因式定理，它是解多项式方程最有力的工具之一。

If (P(a)=0), then (x=a) satisfies the equation, so (x-a) divides (P(x)) exactly. Conversely, if (x-a) is a factor, then substituting (x=a) gives a product of factors including (a-a=0), so the value of (P(a)) is 0.

如果 (P(a)=0)，那么 (x=a) 满足方程，所以 (x-a) 能整除 (P(x))。反过来，如果 (x-a) 是因式，那么代入 (x=a) 后，乘积中包含 (a-a=0)，所以 (P(a)=0)。

Worked example: Show that (x=2) is a root of (x^3 – 7x + 6 = 0), and factorise the cubic.

示例：证明 (x=2) 是方程 (x^3 – 7x + 6 = 0) 的一个根，并将该三次式因式分解。

We substitute (x=2) into (P(x)=x^3 – 7x + 6). Since (2^3 – 7(2) + 6 = 8 – 14 + 6 = 0), it follows from the Factor Theorem that ((x-2)) is a factor.

把 (x=2) 代入 (P(x)=x^3 – 7x + 6)。因为 (2^3 – 7(2) + 6 = 8 – 14 + 6 = 0)，所以根据因式定理，((x-2)) 是一个因式。

Dividing (x^3 – 7x + 6) by (x-2) gives (x^2 + 2x – 3). This quadratic factors as ((x+3)(x-1)). Hence the complete factorisation is ((x-2)(x+3)(x-1)).

用 (x^3 – 7x + 6) 除以 (x-2)，得到 (x^2 + 2x – 3)。这个二次式可以分解为 ((x+3)(x-1))。因此完整的因式分解为 ((x-2)(x+3)(x-1))。

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2. The Remainder Theorem | 余数定理

When a polynomial (P(x)) is divided by a linear expression ((x-a)), the remainder is exactly (P(a)). This is the Remainder Theorem. It allows you to find remainders without performing the full polynomial division.

当多项式 (P(x)) 除以一次式 ((x-a)) 时，余数恰好等于 (P(a))。这就是余数定理。它使你不必做完整的多项式除法就能求出余数。

The Remainder Theorem is a special case of the division rule (P(x)=(x-a)Q(x)+R), where (Q(x)) is the quotient and (R) is a constant. Substituting (x=a) immediately gives (P(a)=R).

余数定理是除法规则 (P(x)=(x-a)Q(x)+R) 的特殊情形，其中 (Q(x)) 是商式，(R) 是常数。把 (x=a) 代入，立即得到 (P(a)=R)。

Example: Find the remainder when (4x^3+2x^2-5x+1) is divided by (x-2).

示例：求 (4x^3+2x^2-5x+1) 除以 (x-2) 的余数。

Let (P(x)=4x^3+2x^2-5x+1). By the Remainder Theorem, the remainder is (P(2)=4(8)+2(4)-10+1=32+8-10+1=31).

令 (P(x)=4x^3+2x^2-5x+1)。根据余数定理，余数为 (P(2)=4(8)+2(4)-10+1=32+8-10+1=31)。

In AQA exam questions, the Remainder Theorem is frequently combined with the Factor Theorem. For example, you may need to find a missing coefficient using both facts.

在 AQA 考试题目中，余数定理经常与因式定理结合使用。例如，你可能需要同时利用这两个结论求出缺失的系数。

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3. Solving Linear and Quadratic Equations | 求解一次与二次方程

Before tackling higher-degree equations, you must be confident with linear and quadratic cases. A linear equation (ax+b=0) has the single solution (x=-dfrac{b}{a}).

在处理高次方程之前，你必须熟练掌握一次和二次方程。一次方程 (ax+b=0) 有唯一解 (x=-dfrac{b}{a})。

For a quadratic equation (ax^2+bx+c=0), you can solve by factorisation, by completing the square, or by using the quadratic formula:

对于二次方程 (ax^2+bx+c=0)，你可以通过因式分解、配方法或二次求根公式来求解：

x = [−b ± √(b² − 4ac)] / (2a)

The expression (b^2-4ac) is called the discriminant. If it is positive, there are two distinct real roots; if it is zero, there is one repeated real root; if it is negative, there are no real roots but two complex conjugate roots.

表达式 (b^2-4ac) 称为判别式。若它为正，则方程有两个不同的实根；若为零，则方程有一个重实根；若为负，则没有实根，而是有一对共轭复数根。

These ideas extend naturally to higher-degree polynomials. A cubic equation (P(x)=0) can be attacked by first finding one root using the Factor Theorem, then reducing the problem to a quadratic.

这些思想自然推广到高次多项式。对于三次方程 (P(x)=0)，可以先用因式定理找到一个根，从而把问题化为二次方程。

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4. Solving Cubic Equations Using the Factor Theorem | 用因式定理求解三次方程

A cubic equation has the standard form (ax^3+bx^2+cx+d=0), where (a neq 0). In A-Level work, you are usually expected to find at least one integer root by testing factors of the constant term.

三次方程的标准形式是 (ax^3+bx^2+cx+d=0)，其中 (a neq 0)。在 A-Level 中，通常要求你先通过测试常数项的因子找到一个整数根。

For example, for (2x^3 – 3x^2 – 11x + 6 = 0), the possible integer roots are factors of 6: (pm1, pm2, pm3, pm6). Testing these gives (P(2)=16-12-22+6=-12), while (P(3)=54-27-33+6=0). So (x=3) is a root.

例如，对于 (2x^3 – 3x^2 – 11x + 6 = 0)，可能的整数根是 6 的因子：(pm1, pm2, pm3, pm6)。测试这些值得到 (P(2)=16-12-22+6=-12)，而 (P(3)=54-27-33+6=0)。所以 (x=3) 是一个根。

Since (x=3) is a root, ((x-3)) is a factor. Dividing the cubic by (x-3) gives (2x^2+3x-2), which factors to ((2x-1)(x+2)). Therefore the solutions are (x=3, x=frac{1}{2}, x=-2).

因为 (x=3) 是一个根，所以 ((x-3)) 是因式。用三次式除以 (x-3)，得到 (2x^2+3x-2)，它可分解为 ((2x-1)(x+2))。因此解为 (x=3, x=frac{1}{2}, x=-2)。

This method is systematic: test a rational number, apply the Factor Theorem, divide, then solve the remaining quadratic. Remember that you may need to do the division using algebraic long division or synthetic division.

这个方法是系统的：测试一个有理数，应用因式定理，做除法，然后解剩下的二次方程。请记住，你可能需要使用综合除法或长除法来完成这里的除法运算。

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5. The Rational Root Theorem | 有理根定理

For a polynomial (a_nx^n + a_{n-1}x^{n-1} + cdots + a_1x + a_0 = 0), the Rational Root Theorem states that any rational root, when written in lowest terms (frac{p}{q}), must have (p) dividing (a_0) and (q) dividing (a_n).

对于多项式 (a_nx^n + a_{n-1}x^{n-1} + cdots + a_1x + a_0 = 0)，有理根定理表明：任何有理根 (frac{p}{q}) 写成最简形式时，(p) 必须整除 (a_0)，(q) 必须整除 (a_n)。

This theorem is extremely useful for higher-degree equations because it gives a finite list of candidates to test. For instance, if (a_n=1) and (a_0=6), the only possible rational roots are integers that divide 6.

这个定理对高次方程非常有用，因为它给出了一份有限的候选名单用于测试。例如，如果 (a_n=1)，(a_0=6)，那么唯一可能的有理根就是整除 6 的整数。

When testing candidates, be efficient. Substitute each candidate into (P(x)). If the value is zero, you have found a root. Once you have reduced the degree to 2, you can finish with factorisation or the quadratic formula.

在测试候选值时，要讲究效率。将每个候选值代入 (P(x))。如果值为零，就找到了一个根。一旦把方程次数降低到 2，就可以用因式分解或二次公式完成求解。

Worked example: Solve (x^4 – 5x^3 + 5x^2 + 5x – 6 = 0). Possible rational roots are (pm1, pm2, pm3, pm6). Testing (x=1): (1-5+5+5-6=0), so (x=1) is a root.

示例：解方程 (x^4 – 5x^3 + 5x^2 + 5x – 6 = 0)。可能的有理根是 (pm1, pm2, pm3, pm6)。测试 (x=1)：(1-5+5+5-6=0)，所以 (x=1) 是一个根。

Dividing by (x-1) yields (x^3 – 4x^2 + x + 6). Testing (x=2): (8-16+2+6=0), so (x=2) is a root of this cubic. Dividing again gives (x^2 – 2x – 3 = (x-3)(x+1)). Therefore the full solution set is (x=1,2,3,-1).

除以 (x-1) 得到 (x^3 – 4x^2 + x + 6)。测试 (x=2)：(8-16+2+6=0)，所以 (x=2) 是这个三次式的根。再次作除法得到 (x^2 – 2x – 3 = (x-3)(x+1))。因此完整解集为 (x=1,2,3,-1)。

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6. Repeated Roots and Multiplicity | 重根与重数

If a polynomial equation has a factor ((x-a)^m), then (x=a) is a root of multiplicity (m). For (m=1), the root is simple; for (m=2), it is a double root; for (m=3), a triple root; and so on.

如果多项式方程含有因式 ((x-a)^m)，那么 (x=a) 是该方程的 (m) 重根。当 (m=1) 时称为单根；(m=2) 时称为二重根；(m=3) 时称为三重根；以此类推。

Geometrically, a repeated root corresponds to a point where the graph of (P(x)) touches the (x)-axis without crossing it. If the multiplicity is odd, the graph crosses; if it is even, the graph touches and turns.

从几何上看，重根对应多项式图像与 (x) 轴相切而不穿过该点的情形。若重数为奇数，图像会穿过 (x) 轴；若为偶数，图像只是相切并折返。

There is also an algebraic connection to the derivative. If (P(a)=0) and (P'(a)=0), then (a) is at least a double root. If (P'(a)=0) and (P”(a)=0) as well, the multiplicity is at least three.

重根还与导数有代数联系。若 (P(a)=0) 且 (P'(a)=0)，那么 (a) 至少是二重根。若同时还有 (P”(a)=0)，则重数至少为三。

Example: For (P(x)=x^3-3x+2), check whether (x=1) is a repeated root. We have (P(1)=1-3+2=0) and (P'(x)=3x^2-3), so (P'(1)=0). Thus (x=1) is a repeated root. Indeed, (x^3-3x+2=(x-1)^2(x+2)).

示例：对于 (P(x)=x^3-3x+2)，判断 (x=1) 是否为重根。我们有 (P(1)=1-3+2=0)，且 (P'(x)=3x^2-3)，所以 (P'(1)=0)。因此 (x=1) 是重根。事实上，(x^3-3x+2=(x-1)^2(x+2))。

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7. Sum and Product of Roots | 根的和与积

For a quadratic equation (ax^2+bx+c=0) with roots (alpha) and (beta), we have the useful identities:

对于根为 (alpha) 和 (beta) 的二次方程 (ax^2+bx+c=0)，我们有下列重要关系：

α + β = −b/a, αβ = c/a

These relationships come from expanding (a(x-alpha)(x-beta)) and comparing coefficients. They allow you to write down the equation from known roots without expanding brackets.

这些关系来自展开 (a(x-alpha)(x-beta)) 并比较系数。它们允许你在已知根的情况下直接写出方程，而不必展开括号。

For a cubic equation (ax^3+bx^2+cx+d=0) with roots (alpha, beta, gamma), the corresponding results are:

对于根为 (alpha, beta, gamma) 的三次方程 (ax^3+bx^2+cx+d=0)，相应结果为：

α+β+γ = −b/a, αβ+βγ+γα = c/a, αβγ = −d/a

For a quartic equation (ax^4+bx^3+cx^2+dx+e=0) with roots (alpha, beta, gamma, delta), the sum of the roots is (−b/a), the sum of products in pairs is (c/a), the sum of products in triples is (−d/a), and the product of all four roots is (e/a).

对于根为 (alpha, beta, gamma, delta) 的四次方程 (ax^4+bx^3+cx^2+dx+e=0)，根之和为 (−b/a)，两两乘积之和为 (c/a)，三三乘积之和为 (-d/a)，四个根的乘积为 (e/a)。

These formulas are often tested in AQA questions where you are given partial information about the roots, such as one root being the reciprocal of another.

在 AQA 试题中，常常会给出关于根的部分信息，例如一个根是另一个根的倒数，这时就要用到这些公式。

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8. Finding Equations from Related Roots | 从相关根构造方程

A common problem asks you to form a new polynomial equation whose roots are related to the roots of a given equation. For instance, if (alpha) and (beta) are roots of (x^2+px+q=0), find the equation whose roots are (2alpha) and (2beta).

一类常见问题是：已知原方程的根 (alpha) 和 (beta)，要求构造一个新方程，使得新方程的根与原方程的根有某种关系。例如，若 (alpha) 和 (beta) 是 (x^2+px+q=0) 的根，求以 (2alpha) 和 (2beta) 为根的方程。

One method is to use the substitution (y=2x). Since (y) equals twice the original root, the new roots in terms of (y) are (y/2). Substituting (x=y/2) into the original equation and simplifying gives the required equation in (y).

一种方法是使用代换 (y=2x)。因为新根 (y) 是原根的两倍，所以用 (y) 表示原来的根为 (y/2)。将 (x=y/2) 代入原方程并化简，就得到关于 (y) 的所需方程。

Example: Given (x^2-3x+2=0) with roots (alpha, beta), find the equation with roots (frac{1}{alpha}) and (frac{1}{beta}).

示例：已知 (x^2-3x+2=0) 的根为 (alpha, beta)，求以 (frac{1}{alpha}) 和 (frac{1}{beta}) 为根的方程。

Here (alpha+beta=3) and (alphabeta=2). The new roots have sum (frac{1}{alpha}+frac{1}{beta} = frac{alpha+beta}{alphabeta} = frac{3}{2}), and product (frac{1}{alphabeta} = frac{1}{2}). Hence the new equation is (x^2 – frac{3}{2}x + frac{1}{2}=0), or equivalently (2x^2-3x+1=0).

这里 (alpha+beta=3)，(alphabeta=2)。新根之和为 (frac{1}{alpha}+frac{1}{beta} = frac{alpha+beta}{alphabeta} = frac{3}{2})，乘积为 (frac{1}{alphabeta} = frac{1}{2})。因此新方程为 (x^2 – frac{3}{2}x + frac{1}{2}=0)，等价于 (2x^2-3x+1=0)。

Another powerful technique is the substitution (y = g(x)) such as (y=x-3) or (y=x^2), which can transform a degree-n equation into a simpler one. Recognising these transformations is a key exam skill.

另一种常用技巧是代换 (y = g(x))，如 (y=x-3) 或 (y=x^2)，它能把一个 (n) 次方程转化为更简单的方程。识别这些变换是一项重要的考试技能。

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9. Solving Quartic and Higher-Degree Equations | 求解四次及更高次方程

Quartic equations can sometimes be reduced to quadratics by rewriting the expression. A particularly nice case is the quartic that contains only even powers of (x): (ax^4+bx^2+c=0). This is called a quadratic in (x^2).

四次方程有时可以通过改写表达式化为二次方程。一个特别好的例子是只含 (x) 的偶次幂的四次方程：(ax^4+bx^2+c=0)。这被称为关于 (x^2) 的二次方程。

To solve it, set (u=x^2). Then (u^2=x^4), so the equation becomes (au^2+bu+c=0). Solve this quadratic for (u), then take square roots: (x = pmsqrt{u}).

解这类方程时，令 (u=x^2)。那么 (u^2=x^4)，于是方程化为 (au^2+bu+c=0)。先解出 (u)，再开平方：(x = pmsqrt{u})。

Example: Solve (x^4 – 5x^2 + 4 = 0). Let (u=x^2). Then (u^2 – 5u + 4 = 0), so ((u-1)(u-4)=0), giving (u=1) or (u=4). Therefore (x=pm1) or (x=pm2).

示例：解方程 (x^4 – 5x^2 + 4 = 0)。令 (u=x^2)，得 (u^2 – 5u + 4 = 0)，所以 ((u-1)(u-4)=0)，即 (u=1) 或 (u=4)。于是 (x=pm1) 或 (x=pm2)。

For general higher-degree equations, you should combine the Rational Root Theorem with algebraic division and repeated application of the Factor Theorem. In AQA exams, you will not be asked to solve general quintic equations analytically; the level typically stops at quartic equations that can be factorised neatly.

对于一般的高次方程，你应该结合有理根定理、代数除法以及反复应用因式定理。在 AQA 考试中，通常不会要求你解析求解一般五次方程；难度通常止步于能够干净地因式分解的四次方程。

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10. Exam Strategies for AQA | AQA 考试策略

Always start by rearranging the polynomial equation so that every term is on one side and the other side is zero. Be careful not to divide by a variable factor such as (x), because that could lose the root (x=0).

解题时，首先把多项式方程整理成所有项都在一边、另一边为零的形式。要小心不要把两边同时除以含变量的因式（如 (x)），因为这样可能丢失 (x=0) 这个根。

When using the Factor Theorem, test small integer values first, especially (x=1) and (x=-1). These are the quickest to evaluate and often appear as roots in exam questions.

在使用因式定理时，先测试小的整数值，尤其是 (x=1) 和 (x=-1)。它们的计算最快，而且在考试题目中经常作为根出现。

For quadratic equations, always check the discriminant before deciding whether complex roots exist. In AQA pure mathematics, you should be ready to write the roots in the form (a pm bi).

对于二次方程，在判断是否存在复数根之前，一定要先计算判别式。在 AQA 纯数学中，你应该能够以 (a pm bi) 的形式写出根。

Sketching the graph of (P(x)) helps you estimate the number and location of real roots. Use the sign of the leading coefficient and the positions of any turning points to confirm your algebraic answers.

画出 (P(x)) 的图像有助于你估算实根的个数和位置。利用首项系数的符号以及任何驻点的位置，可以核查你的代数答案。

Finally, write all roots clearly and distinguish between real and complex solutions. In exam answers, always show substitution into the polynomial or the result of division, because method marks are often awarded even when the final answers are wrong.

最后，要清楚地写出所有根，并区分实根和复数根。在考试答案中，务必展示代入多项式或除法过程，因为即使最终答案错误，方法分也常常会被给出。

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Conclusion | 总结

Polynomial equations of degree (n) form a central topic in AQA A-Level Mathematics. Success depends on a firm grasp of the Factor Theorem, Remainder Theorem, rational root testing, and the relationships between roots and coefficients.Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com

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